%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% %% %% On Various Modes of %% Scalar Convergence in L_0(X) %% %% %% S.J. Dilworth & Maria Girardi %% %% %% 5 April 1996 %% %% AMS-TEX version 2.1 %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentstyle{amsppt} %\magnification=\magstep1 \magnification=1200 %% for msri format %\NoRunningHeads \TagsOnRight \hcorrection{.25in} %%%%%%%%%%%%%%%%%%%%%%%%%%% MACROS %%%%%%%%%%%%%%%%%%%% \def\X{\frak X} \def\Y{\frak Y} \def\lor{L_1 (\Bbb R)} \def\lmr{L_0 (\Bbb R)} \def\lpr{L_p (\Bbb R)} \def\lox{L_1 (\X)} \def\lmx{L_0 (\X)} \def\lqx{L_q (\X)} \def\lpx{L_p (\X)} \def\lrx{L_r (\X)} \def\lrr{L_r (\Bbb R)} \def\e{\varepsilon} \def\w{\omega} \def\a{\alpha} \def\D{\Delta} \def\O{\Omega} \def\rn{\right\Vert} \def\ln{\left\Vert} \def\lav{\left\vert} \def\rav{\right\vert} \def\qed{\hfill{\vrule height6pt width6pt depth0pt}} \font\cmcsc=cmcsc8 \def\hf{\hfill} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \def\ra{\longrightarrow} \def\la{\longleftarrow} \def\ua{\uparrow} \def\da{\downarrow} \def\ane{\nearrow} \def\asw{\swarrow} \def\ase{\searrow} \def\anw{\nwarrow} \def\nra{\longrightarrow\hskip-.9em\shortmid} \def\nla{\longleftarrow\hskip-.9em\shortmid} \def\nua{\uparrow\hskip-.35em\nshortmid} \def\nda{\downarrow\hskip-.35em\nshortmid} \def\nane{\nearrow\hskip-.58em\shortmid} \def\nasw{\swarrow\hskip-.565em\shortmid} \def\nase{\searrow\hskip-.54em\shortmid} \def\nanw{\nwarrow\hskip-.54em\shortmid} \def\rala{^{\longrightarrow}_{\longleftarrow}} \def\ranla{^{\longrightarrow}_{\nla}} \def\span{\text{sp\,}} \hyphenation{sca-lar-ly} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \topmatter %%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \title On various modes of Scalar Convergence in $\lmx$ \endtitle \author S. J. Dilworth \\ Maria Girardi \endauthor \thanks Both authors participated in the NSF Workshop in Linear Analysis \& Probability, Texas A\&M University (supported in part by NSF grant DMS-9311902), and in the Convex Geometry and Geometric Functional Analysis Program at the Mathematical Sciences Research Institute, Berkeley, CA (supported in part by NSF grant DMS-9022140). Girardi is supported in part by NSF grant DMS-9306460. \endthanks \address \hfill\newline Department of Mathematics, University of South Carolina, Columbia, SC 29208, U.S.A. \endaddress \email dilworth\@math.sc.edu, girardi\@math.sc.edu \endemail \subjclass 28A20, 46E30, %spaces of measurable functions 46E40. %spaces of vector- and operator-valued functions \endgraf {\it Date:} 5 April 1996 \endsubjclass \abstract A sequence $\{f_n\}$ of strongly-measurable functions taking values in a Banach space $\X$ is scalarly null a\.e\. (resp. scalarly null in measure) if $x^*f_n \rightarrow0$ a\.e\. (resp. $x^*f_n \rightarrow 0$ in measure) for every $x^*\in \X^*$. Let $1\le p\le \infty$. The main questions addressed in this paper are whether an $L_p(\X)$-bounded sequence that is scalarly null a\.e\. will converge weakly a\.e\. (or have a subsequence which converges weakly a\.e\.), and whether an $L_p(\X)$-bounded sequence that is scalarly null in measure will have a subsequence that is scalarly null a\.e. The answers to these and other similar questions often depend upon $p$ and upon the geometry of $\X$. \endabstract %\headline={\ifodd\pageno\rightheadline\else\leftheadline\fi} %\def\rightheadline{\eightrm\hfil Scalar Convergence \hfil\folio} %\def\leftheadline{\eightrm\hfil DILWORTH \ and \ GIRARDI \hfil\folio} \endtopmatter \document \heading{1. Introduction}\endheading Consider the space $\lmx$ of strongly-measurable functions defined on the usual % \linebreak Lebesgue measure space $(\Omega, \Sigma, \mu)$ on $[0,1]$ and taking values in the Banach space~ $\X$. Among the most important linear subspaces of $L_0(\X)$ are the Bochner-Lebesgue spaces $\lpx$ for $1\leq p \leq \infty$. When $\X = \Bbb R$, %and $p = 0$ or $1 \leq p \leq \infty$ (or $p=0$), the $L_p(\X)$ spaces are just the usual Lebesgue spaces, which we shall denote by $L_p$. A sequence $\{ f_n \}$ in $\lmx$ may converge to $f$ in $\lmx$ in a variety of modes. In this paper, we examine the implications going between the four modes described below. The sequence $\{ f_n \}$ converges {\it scalarly a\.e\.\/} (resp.~ {\it scalarly in $L_p$\/} for a fixed $1 \leq p \leq \infty$, resp.~ {\it scalarly in measure\/}) to $f$ if for each $x^*$ in the dual space ~$\X^*$ of~ $\X$, the corresponding sequence $\{ x^* f_n \}$ in $\lmr$ converges almost everywhere (resp.~ in $\lpr$, resp.~ in measure) to $x^*f$. Thus, $\{ f_n \}$ converges scalarly a\.e\. to $f$ if for each $x^*$ in $\X^*$ there is a set $A$ (which depends on $x^*$) of full measure such that $\{ x^* f_n (\omega) \}$ converges to $x^* f(\omega)$ for each $\omega$ in $ A$. If the sequence satisfies the stronger property that there is a set $A$ (independent of $x^*$) of full measure such that $\{ x^* f_n (\omega) \}$ converges to $x^* f(\omega)$ for each $\omega$ in $A$ and each $x^*$ in $\X^*$, then we say that $\{ f_n \}$ converges {\it weakly almost everywhere\/} (i\.e\.~{\it weakly a\.e\.\/}) to $f$. The following obvious positive and negative implications hold: $$\vbox{ \settabs\+\indent &\hskip 1 true in &\hskip 2 true in &\hskip 1 true in& \cr \+&\hf weakly a\.e\.\hf&\hf$\ra$\hf&\hf scalarly a\.e\. \hf\cr \+\cr \+&\hf$\nua~\nda$\hf&\hf$\nane~\nasw\quad\ase~\nanw$\hf&\hf$\nua~\da$\hf&\cr \+\cr \+&\hf scalarly in $L_p$ \hf&\hf$\ranla$\hf&\hf scalarly in measure \cr}$$ for each $1\leq p < \infty$. Similarly: $$\vbox{ \settabs\+\indent &\hskip 1 true in &\hskip 2 true in &\hskip 1 true in& \cr \+&\hf weakly a\.e\.\hf&\hf$\ra$\hf&\hf scalarly a\.e\. \hf\cr \+\cr \+&\hf$~\nda$\hf&\hf$\ane~\nasw\quad\ase~\nanw$\hf&\hf$\nua~\da$\hf&\cr \+\cr \+&\hf scalarly in $L_p$ \hf&\hf$\ranla$\hf&\hf scalarly in measure \cr}$$ for $p = \infty$. Sets of the form $$ V_{x^*, \e}(0) = \left\{ g\in L_0(\X) \: \mu \left\{ \lav x^* (g) \rav \geq \e \right\} < \e \right\} \ , $$ where $x^* \in \X^*$ and $\e >0$, form a local subbasis at zero for the translation-invariant topology of scalar convergence in measure. Endowed with this topology, $L_0(\X)$ is a non-locally convex Hausdorff topological vector space. For a fixed $1\leq p \leq \infty$, sets of the form $$ S_{x^* , \e}(0) = \left\{ g \in L_p (\X) \: \ln x^* g \rn_{L_p} < \e \right\} \ , $$ where $x^* \in \X^*$ and $\e>0$, form a local subbasis at zero for the translation-invariant topology of scalar convergence in $L_p$. Endowed with this topology, $L_p(\X)$ is a locally convex Hausdorff topological vector space. Let $\X_0$ be a norm-closed subspace of $\X$. The Hahn-Banach Theorem quickly gives two observations. First, a sequence of $L_0(\X_0)$ functions converges to the null function in one of the above modes when viewed as a sequence in $L_0(\X_0)$ if and only if it does so when viewed as a sequence in $L_0(\X)$. Secondly, the topology of scalar convergence in measure on $L_0(\X_0)$ coincides with the subspace topology inherited from the topology of scalar convergence in measure on $L_0(\X)$. Let us show that under this identification $L_0(\X_0)$ is in fact a {\it closed} subspace of $L_0(\X)$. \proclaim{Proposition 1.1} Let $\X_0$ be a (norm-closed) subspace of $\X$. Then $L_0(\X_0)$ is a closed subspace of $L_0(\X)$ in the topology of scalar convergence in measure. In particular, if $\{f_n\}$ is a sequence of $\X_0$-valued functions in $L_0(\X)$ that converges scalarly in measure to $f$ in $ L_0(\X)$, then $f$ is also $\X_0$-valued. \endproclaim \demo{Proof} Let $f\in L_0(\X)$ belong to the closure of $L_0(\X_0)$ in the topology of scalar convergence in measure. Since the range of $f$ is essentially separably-valued, there is a subset $Y \supset \X_0$ such that $Y/\X_0$ is separable and $f(\omega) \in Y$ a\.e\. By the Hahn-Banach theorem there exists a sequence $\{x_n^*\}$ in $\X^*$ such that if $y \in Y$ then $y \in \X_0$ if and only if $x_n^*(y)=0$ for all $n \ge 1$. In particular, if $g \in L_0(\X_0)$ then $x_n^*(g(\omega))=0$ for all $n \ge1$ a\.e\., and it follows that $x_n^*(f(\omega))=0$ for all $n \ge 1$ a\.e\., which proves that $f \in L_0(\X_0)$. \qed \enddemo For $1 \leq p \leq \infty$, it is easy to see that $L_p (\X)$ is {\it not} a closed subspace of $L_0 (\X)$ in the topology of scalar convergence in measure for any Banach space $\X$. So we consider the unit ball $$B(L_p(\X)) = \{ g\in L_0(\X) \: \ln g \rn_{L_p(\X)} \leq 1 \}$$ of $L_p(\X)$. \proclaim{Proposition 1.2} For $1\le p\le \infty$, the unit ball $B(L_p(\X))$ of $L_p(\X)$ is closed in $L_0(\X)$ in the topology of scalar convergence in measure. \endproclaim \demo{Proof} Consider $f \in L_0(\X)\setminus B(L_p(\X))$. It is sufficient to find an open neighborhood about $f$ that does not meet $ B(L_p(\X))$. Fix $\e > 0$ so that $(1-4\e) \ln f \rn_{L_p(\X)} > 1$. Since $f$ is strongly-measurable, there is a countably-valued function $g\in L_0(\X)$ satisfying $$\ln f(\omega) - g(\omega) \rn_{\X}~\leq~ \e~ \max\{\ln f(\omega)\rn_{\X},\ln g(\omega)\rn_{\X}\}$$ for almost all $\omega$. By making an appropriate choice of representative we may write $g=\sum_k x_k 1_{E_k}$, where $\{E_k\}_k$ partitions the support of $g$ into sets of strictly positive measure. Now $\ln g \rn_{L_p(\X)} \ge (1-\e) \ln f \rn_{L_p(\X)}$, and so $(1 -3\e)\ln g \rn_{L_p(\X)}>1$. Hence we may choose $N \in \Bbb N$ so that $$(1 -3\e)\ln \tilde g \rn_{L_p(\X)}>1~,$$ where $\tilde g= \sum_{k=1}^N x_k 1_{E_k}$. Now find $\{ x^*_k \}_{k=1}^N$ in $S(\X^*)$ with $x^*_k (x_k) = \ln x_k \rn $. Consider the following neighborhood of $f$ in the topology of scalar convergence in measure: $$U= \bigcap_{k=1}^N \left\{ h\in L_0(\X) \: \mu \left\{ \lav x^*_k (h-f) \rav \geq \e \ln x_k \rn \right\} < \e \mu(E_k) \right\}~. $$ For $1 \leq k \leq N$, if $\omega \in E_k$ then % $\tilde g(\omega) = \|x_k\|$ and also $\|f(\omega) - \tilde g(\omega)\| \le \e \| x_k \|$; thus, for each $h \in U$ $$ \mu \{ \omega \in E_k: \|h(\omega)\| \ge (1 - 2\e) \| x_k \| \} \ge (1-\e)\mu(E_k)~. $$ Hence $$ \ln h \rn_{L_p(\X)} \ge (1-\e)^{1/p}(1-2\e) \ln \tilde g \rn_{L_p(\X)}>(1-3\e)\ln \tilde g \rn_{L_p(\X)}>1~, $$ following the convention that $1/\infty = 0$. So $U$ does not intersect $ B(L_p(\X))$, as required. \qed \enddemo This suggests imposing the following natural boundedness conditions: a sequence $\{ f_n \}$ of $L_p(\X)$ functions is said to be: \roster \item"{-}" {\it pointwise bounded a\.e\.} if $\sup_n \ln f_n (\omega) \rn_{\X}<\infty$ for each $\omega$ in some set of full measure. \item"{-}" bounded in $L_p(\X)$ (for short, {\it $L_p(\X)$-bounded}) if $\sup_n \ln f_n \rn_{L_p(\X)}<\infty$. \endroster From Proposition 1.2, we see that sequences of the latter type are well-behaved in the following sense. \proclaim{Corollary 1.3} Let $1\le p\le\infty$ and suppose that $\{f_n\}$ is bounded in $L_p(\X)$. If $\{f_n\}$ converges scalarly in measure to $f\in L_0(\X)$, then $f\in L_p(\X)$. \endproclaim \demo{Remarks} 1. Hence, in discussing the convergence (in any one of the above four modes) of a sequence in $L_0(\X)$ of functions valued in a subspace $\X_0$ of $\X$, there is no loss of generality in taking the limit function to be the null function and viewing the sequence as in $L_0(\X_0)$. \newline 2. Similarly, if we choose to restrict ourselves to the subset $L_p(\X)$ of $L_0(\X)$, in discussing scalar convergence in measure for an $L_p(\X)$-bounded sequence, there will be no loss of generality in taking the limit function to be the null function. \newline 3. The question of the {\it existence} of a limit for an $L_p(\X)$-bounded {\it Cauchy sequence} in the topology of scalar convergence of measure is more problematic and will be deferred until Section~5. \enddemo We will also use the following elementary facts without further comment. Fact~1.4 provides a useful necessary condition for weak a\.e\. convergence while Fact~1.5 will be used to prove scalar convergence a\.e\. \demo{Fact 1.4} A weakly convergent sequence in a Banach space is norm-bounded. Thus, %Hence, if for a given sequence $\{ f_n \}$ in $L_0(\X)$, there exists a subset $B$ of strictly positive $\mu$-measure such that $\limsup \ln f_n (\omega) \rn = \infty$ for each $\omega \in B$, then $\{ f_n \}$ does not converge weakly a\.e\. \enddemo \demo{Fact 1.5} A sequence $\{ f_n \}$ in $L_1$ converges to the null function a.e. whenever $\sum \ln f_n \rn_{L_1} < \infty$. \enddemo If $Y$ is a subset of $\X$, then $\span Y $ denotes the linear span of $Y$ and $\left[ Y \right]$ denotes the closed linear span of $Y$. All notation and terminology, not otherwise explained, are as in [DU] or [LiT]. \heading{2. Convergence Properties}\endheading Proposition~1.2 suggests that it would be of interest to study the following properties that a Banach space $\X$ might enjoy. \roster \item"($A_p^\prime$)" Each $L_p(\X)$-bounded sequence of functions that converges scalarly a\.e\. also converges weakly a.e. \item"($A_p$)" Each $L_p(\X)$-bounded sequence of functions that converges scalarly a\.e\. has a subsequence that converges weakly a.e. \item "($B_p$)" Each $L_p(\X)$-bounded sequence of functions that converges scalarly in measure has a subsequence that converges weakly a.e. \item"($C_p$)" Each $L_p(\X)$-bounded sequence of functions that converges scalarly in $L_p$ has a subsequence that converges weakly a.e. \item"($D_p$)" Each $L_p(\X)$-bounded sequence of functions that converges scalarly in measure has a subsequence that converges scalarly a.e. \item"($E_p)$" Each $L_p(\X)$-bounded sequence of functions that converges scalarly in $L_p$ has a subsequence that converges scalarly a.e. \endroster For convenience, a schematic summary of the properties is given below, in which a double arrow indicates an implication that is always valid. $$ \vbox{ \settabs\+indent &\hskip 75 pt &\hskip 40 pt &\hskip 40 pt &\hskip 60 pt\cr \+&weakly a\.e\. &$\Rightarrow$ &$\longleftarrow^{\hskip-1 em A}$ &scalarly a\.e\.\cr \+& \cr \+&\hskip 9 pt{\smc c}$\uparrow$ &{\smc e}$\nearrow$ &$\nwarrow${\smc b} &\hskip 9 pt $\uparrow${\smc d} \hskip 11 pt $\Downarrow$ \cr \+& \cr \+&scalarly in $L_p$ &$\Rightarrow$ & &scalarly in measure \quad . \cr } $$ \demo{Remarks} 1. Clearly, subsequential convergence is the most one can expect in passing from scalar in measure or scalar in $L_p$ to scalar a.e. convergence.\newline 2. Note that if $\X$ has (Property$_p$) and $p < q$, then $\X$ also has (Property$_q$). \newline 3. Note the following obvious implications. $$\vbox{ \settabs\+\indent&\hskip 30 pt &\hskip 30 pt &\hskip 30 pt &\hskip 30 pt &\hskip 30 pt\cr \+&($A_p$)&$\longleftarrow$&($B_p$)&$\longrightarrow$&($C_p$)\cr \+& & &\hskip 7 pt$\downarrow$ & &\hskip 7 pt$\downarrow$\cr \+& & &($D_p$)&$\longrightarrow$&($E_p$)\cr } $$ \newline 4. For a fixed $1 \leq p < \infty$, a sequence $\{f_n\}$ in $L_0(\X)$ converges scalarly in $L_p$ if and only if (i) $\{f_n\}$ converges scalarly in measure and (ii) for each $x^* \in \X^*$, the set $\{ \lav x^* f \rav^p \}$ is uniformly integrable. Note that (ii) holds when $\{f_n\}$ is $L_r(\X)$-bounded for some $r > p$. \enddemo \heading{3. $L_\infty(\X)$-Bounded Sequences}\endheading \subhead [3.i] From scalar convergence to weak a.e. convergence \endsubhead In this subsection, we characterize those spaces having ($A_\infty$), ($B_\infty$), and ($C_\infty$). \proclaim{Theorem~3.1} Let $\X^*$ have the Radon-Nikod\'ym property. Then $\X$ enjoys the following properties: \roster \item Let $\{f_n\}$ be a sequence in $L_0(\X)$ which converges scalarly a\.e\. to some $f \in L_0(\X)$. Then $\{f_n\}$ converges weakly a\.e\. to $f$ if and only if $\{f_n\}$ is pointwise-bounded a\.e\. \item Let $\{f_n\}$ be a sequence in $L_0(\X)$ which converges scalarly a\.e\. to some $f \in L_0(\X)$. Then $\{f_n\}$ has a subsequence which converges weakly a\.e\. to $f$ if and only if $\{f_n\}$ has a subsequence which is pointwise-bounded a\.e\. \item Let $\{f_n\}$ be a sequence in $L_0(\X)$ which converges scalarly in measure. Then $\{f_n\}$ has a subsequence which converges weakly a\.e\. if and only if $\{f_n\}$ has a subsequence which is pointwise-bounded a\.e\. \endroster \endproclaim \demo{Proof} By Fact 1\.4 each a\.e\. weakly convergent sequence is bounded a\.e\., and so necessity in (1)-(3) (for an {\it arbitrary} Banach space) is clear. To prove sufficiency observe that we may take $f=0$ without loss of generality. We may also assume that $\X^*$ is separable. (Indeed, by the Pettis measurability theorem [P], there is a separable subspace $\X_0$ of $\X$ such that the $f_n$'s are essentially valued in $\X_0$. Because $\X^*$ has the RNP, $\X^*_0$ must be separable [S, Thm.~2].) Let $\{ x_i^* \}$ be dense in $\X^*$. Now we prove sufficiency in (1). Since $\{ f_n \}$ is scalarly null a\.e\., it follows that for each $i$ there is a set $A_i$ of full measure such that if $\omega\in A_i$, then $\lim_n x^*_i f_n(\omega) = 0$. Put $A = \cap_i A_i$. Since the $f_n$'s are pointwise-bounded on some set $B$ of full measure, and since $\{ x_i^* \}$ is dense in $\X^*$, it follows that $\lim _n x^* f_n (\omega) = 0 $ for each $x^* \in \X^*$ and for each $\omega\in A\cap B$. Thus, $\{ f_n \}$ is weakly null a\.e\. Sufficiency in (2) follows at once. Finally, we prove sufficiency in (3). By first passing to a subsequence we may assume that $\{f_n\}$ is pointwise bounded almost everywhere. Since $\{ f_n \}$ converges to zero scalarly in measure, for each $i$ the sequence $\{ x^*_i f_{n} \}_n$ converges in measure to the null function. So by a Cantor diagonalization argument there exists a subsequence $\{ f_{n_k} \}$ such that for almost all $\omega$ $$ \lim_k x^*_i \left( f_{n_{k}} \left(\omega\right)\right) = 0 \ . $$ for all $i$. Now, arguing as before, the pointwise boundedness implies that $\{ f_{n_k} \}$ is weakly null a\.e\. \qed \enddemo For the Banach spaces $\ell_1$, $C(\Delta)$, and the James tree space, Davis and Johnson [DJ] constructed examples of $L_\infty(\X)$-bounded sequences that converge scalarly a\.e\. but not weakly a.e. They conjectured that such a sequence exists for any space $\X$ whose dual fails the Radon-Nikod\'ym property (RNP). Combined with work of Uhl [U] and Stegall [S], a result of Edgar [E] shows that their conjecture was correct. In fact, rather more can be said as the following theorem (whose proof was inspired by [E]) shows. \proclaim{Theorem 3.2} For a Banach space $\X$, the following are equivalent: \roster \item $\X^*$ has the Radon-Nikod\'ym property (i\.e\. $\X$ is an Asplund space); \item $\X$ has ($A_\infty^\prime$); \item $\X$ has ($A_\infty$); \item $\X$ has ($B_\infty$). \endroster \endproclaim \flushpar \demo{Proof} Several implications follow from Theorem~3.1. To prove the other implications, suppose that $\X^*$ fails the RNP. Then [U] there is a separable subspace $\X_0$ of $\X$ such that $\X_0^*$ is not separable. We shall construct an $L_\infty(\X)$-bounded sequence $\{ g_n \}$ of $\X_0$-valued functions such that $g_n \to 0$ scalarly a\.e\. and scalarly in $\lrx$ for $1 \leq r < \infty$, but such that no subsequence of $\{g_n\}$ converges weakly a\.e. (This particular construction has been fruitful in several similar characterizations of $\X^*$ having the RNP [e\.g\.~E, GS, DG].) This will show that $\X$ fails ($A_\infty$), thus completing the proof of the theorem. Let $\Delta = \{ -1, 1 \}^{\Bbb N}$ be the Cantor group with Haar measure $\nu$. Let $\{ \Delta^n_k \: k=1,\ldots, 2^n \}$ be the standard $n$-th partition of $\D$. Thus $\D^0_1 = \D$ and $\D^n_k = \D^{n+1}_{2k-1} \cup \D^{n+1}_{2k}$ and $\nu(\D^n_k) = 2^{-n}$. Instead of our usual Lebesgue measure space on $[0,1]$, we shall now take our underlying measure space to be the completion of $\nu$ for the completion of the Borel $\sigma$-algebra of $\Delta$. Thus $L_p(\X)$ will denote $L_p(\D,\nu;\X)$. We consider the space $C(\D)$ of real-valued continuous functions on $\D$ as a subspace of $L_\infty (\D,\nu)$. Let $ \{ 1_{\D} \} \cup \{ h^n_k \: n=0,1,2, \ldots \text{ and } k=1,\ldots, 2^n \}$ be the usual Haar basis of $C(\D)$, where $h^n_k \: \D \to \Bbb R$ is given by $$ h^n_k ~=~ 1_{\D^{n+1}_{2k-1}} ~-~ 1_{\D^{n+1}_{2k}} \ . $$ Let $\{ e^n_k \: n=0,1,2, \ldots \text{ and } k=1,\ldots, 2^n \}$ be an enumeration (lexicographically) of the usual $\ell_1$ basis and let $H \: \ell_1 \to L_\infty$ be the Haar operator that takes $e^n_k$ to $h^n_k$. By Stegall's Factorization Theorem [S, Theorem~4], $H$ factors through $\X_0$, i\.e\. there are bounded linear operators $R \: \ell_1 \to \X_0 $ and $S \: \X_0 \to L_\infty$ such that $ H = SR$. $$\vbox{ \settabs\+\indent &\hskip 1 true in &\hskip 1 true in &\hskip 1 true in& \cr \+&\hf$\ell_1\quad$ &\hf$\longrightarrow^{^{\hskip-1.25em H}}\hskip1em$\hf &$L_\infty(\D)\supset C(\D)$\hf&\cr \+\cr \+&\hf$\searrow_{^{\hskip-1.3em R}}$& &$\nearrow_{^{\hskip-.1em S}}$\hf&\cr \+\cr \+& &\hf$\X_0$\hf&\cr }$$ Let $\tilde R$ be the natural extension of $R$ to a bounded linear operator from $L_1(\ell_1)$ to $L_1(\X_0)$. Consider the sequence $\{ f_m \}$ of $L_1(\ell_1)$ functions given by $$ f_m (\cdot) ~=~ \frac{1}{m} \sum_{n=1}^m \sum_{k=1}^{2^n} h^n_k(\cdot) e^n_k \ . $$ Let $g_m = \tilde R (f_m)$. Clearly, $\{ g_m \}$ is $L_\infty(\X)$-bounded since $|| f_m ( \w ) ||_{\ell_1} = 1$ for $\nu$-a.e. $\w$ and each $m$. To examine the scalar behavior of $\{ g_m \}$, note that if $y^*\in \X_0^*$, then $$y^* g_m(\cdot) = (R^* y^*) f_m (\cdot)~,$$ where $R^* y^* \in \ell^*_1$. So to show that $\{ g_m \}$ converges to the null function scalarly a\.e\. and scalarly in $L_r (\X)$ we need only show the same for $\{ f_m \}$. So fix a functional $x^*$ in $\ell^*_1$; let $x^*$ have the form $(\alpha^n_k) \in \ell_\infty$, lexicographically ordered. Then $$ x^* f_m(\omega) = \frac{1}{m} \sum_{n=1}^m X_n(\cdot) \text{\qquad where\qquad} X_n(\cdot) = \sum_{k=1}^{2^n} h^n_k(\cdot) \alpha^n_k \ . $$ Note that $\|X_n\|_\infty \le \|x^*\|$, that each $X_n$ has zero mean, and that $\int X_n X_m \, d\nu = 0$ when $n\neq m$. The Strong Law of Large Numbers for uncorrelated random variables with uniformly bounded second moments [cf.~C, Thm.~5.1.2] gives that $\{ x^* f_m \}$ converges to the null function a\.e. Since $\|x^*f_m\|_\infty\le\|x^*\|$ it also follows that $\{x^*f_m\}$ converges to the null function in $L_p$ for $0\le p<\infty$. We shall now show that no subsequence of $\{g_m\}$ converges weakly a\.e\. Since $\{g_m\}$ is scalarly null a\.e\., it suffices to show that no subsequence is weakly null a\.e\. For $\omega\in\Omega$, let $\ell_\omega\in\left[ C\left(\Delta\right)\right]^*$ be the point evaluation at $\omega$ functional and let $\widetilde \ell_\omega\in\left[ L_\infty\left(\Delta\right)\right]^*$ be any Hahn-Banach extension. Then $$ \left(S^*\widetilde\ell_\omega\right) \left( g_n \left(\omega\right)\right) = \ell_\omega \left( H f_n \left(\omega\right)\right)= \left( \frac{1}{m} \sum_{n=1}^m \sum_{k=1}^{2^n} h^n_k(\omega) h^n_k(\cdot)\right) (\omega) =1~, $$ and thus no subsequence of $\{g_n\}$ converges weakly a\.e\. (to the null function) in $L_0(\X_0)$. \qed \enddemo Property $C_\infty$, on the other hand, is a much weaker property according to the following mildly surprising result. \proclaim{Proposition 3.3} Let $\{f_n\}$ be a sequence in $L_0(\X)$ that converges to the null function scalarly in $L_\infty$. Then $\{f_n\}$ is weakly null a.e. In particular, every Banach space enjoys ($C_\infty$) (and {\it a fortiori} ($E_\infty$)). \endproclaim \demo{Proof} For each $n\ge1$, we may write $f_n = g_n + h_n$, where $h_n$ has $L_\infty (\X)$-norm at most $1/n$ and $g_n$ is countably-valued (see e\.g\. [~DU II.1.3]). By choosing a suitable representative of $g_n$ in $L_\infty(\X)$, we may express $g_n$ as $$ g_n = \sum_{k = 1}^\infty x^n_k ~1_{E^n_k}, $$ where each $E^n_k$ has strictly positive measure and, for each $n$, $\Omega$ is the disjoint union of $\{ E^n_k \}_k$. Note that $\{ g_n \}$ also converges to the null function scalarly in $L_\infty (\X)$. Hence, for each $\omega\in\Omega$ and each $x^*\in X^*$, we have $$|x^*(g_n(\omega))|\le \sup_k |x^*(x^n_k)|=\|x^*(g_n)\|_\infty \rightarrow 0$$ as $n\rightarrow \infty$. Hence $\{g_n(\omega)\}$ is weakly null for all $\omega\in\Omega$. Clearly, $\{h_n(\omega)\}$ is norm-null a.e., whence $\{f_n\}$ is weakly null a.e. \qed \enddemo Perhaps the following theorem is the most useful analogue for scalar convergence in general Banach spaces of the familiar fact from real analysis that each sequence that converges in measure has a subsequence that converges a.e. \proclaim{Theorem 3.4} Let $K$ be a weakly compact subset of a Banach space $\X$ and let $\{f_n\}$ be a sequence in $L_0(\X)$ such that each $f_n$ is essentially $K$-valued. If $\{f_n\}$ converges scalarly in measure to $f\in L_0(\X)$, then some subsequence $\{f_{n_k}\}$ converges weakly a\.e\. to $f$. (In particular, $f$ is essentially $K$-valued.) \endproclaim \demo{Proof} Clearly, we may assume that $\X$ is separable. First, we show that $f(\omega)\in K'=\overline{\text{conv}(K)}$ a.e. Suppose this is not the case. Then there exists a closed ball $B\subset \X\setminus K'$ such that $\mu(A)>0$, where $A=\{\omega:f(\omega)\in B\}>0$. By the Hahn-Banach Theorem there exists $x^*\in\X^*$ and $\alpha,\beta \in\Bbb R$ such that $$\sup_{k\in K'}x^*(k)<\alpha<\beta<\inf_{b\in B}x^*(b)~.$$ Since each $f_n$ is (without loss of generality) $K$-valued, it follows that $$ x^*(f_n(\omega))<\alpha<\beta0$. There exists $\delta > 0$ and $N\ge1$ such that $$\{k\in K:|x^*(k)|<\varepsilon\}\supset \cap_{i=1}^N\{k\in K:|x_i^*(k)|<\delta\}~.$$ It follows that $$ \{\omega\in \Omega:|x^*(f_{n_k}(\omega))|<\varepsilon\quad \text{for all sufficiently large k}\}\supset \Omega'~,$$ and so $x^*(f_{n_k}(\omega))\rightarrow 0$ as $n\rightarrow \infty$ for all $\omega\in \Omega'$. Hence $\{f_{n_k}\}$ is weakly null a.e. \qed \enddemo Minor variations in the above proof gives the following result. \proclaim{Theorem 3.5} Let $K$ be a weakly compact subset of a Banach space $\X$ and let $\{f_n\}$ be a sequence in $L_0(\X)$ such that each $f_n$ is essentially $K$-valued. If $\{f_n\}$ converges scalarly a.e. to $f\in L_0(\X)$, then $\{f_n\}$ converges weakly a.e. to $f$. \endproclaim \subhead [3.ii] From scalar convergence in measure to scalar a.e. convergence \endsubhead In the previous subsection, Proposition~3.3 shows that each Banach space enjoys ($E_\infty$). In this subsection, we explore ($D_\infty$). For each $1 \leq p \leq \infty$, it follows directly from the definitions that ($B_p$) implies ($D_p$); however, they are not equivalent. Indeed, Theorem~3.2 implies that $\ell_1$ fails ($B_p$) for each $1 \leq p \leq \infty$. However, $\ell_1$ has ($ D_p$) for each $1 \leq p \leq \infty$, as the next theorem, which was pointed out to us by W.B.~Johnson [cf.~WBJ], shows. We are grateful to him for permission to include this result here. \proclaim{Theorem 3.6} A scalarly null in measure sequence of $L_0(\ell_1)$ functions contains a scalarly null a.e.~subsequence. In particular, $\ell_1$ satisfies properties ($D_p$) for each $1\leq p \leq \infty$. \endproclaim The following lemma will be used in the proofs of several results, including Theorem 3.6. \proclaim{Lemma 3.7} Let $\X$ be a Banach space with a basis and $\{ f_n \}$ be a scalarly in measure null sequence of $\lmx$ functions. There exists a subsequence $\{ f_{n_k} \}$ of $\{ f_n \}$, a blocking $\{ \X_k \}$ of the basis, and sequences $\{ g_k\}$ and $\{h_k\}$ of $\lmx$ functions so that: \roster \item"{(i)}" $\{h_k\}$ converges a.e (in $\X$-norm) to the null function; \endroster and for each $k$: \roster \item"{(ii)}" $f_{n_k} = g_k + h_k$; \item"{(iii)}" $g_k = P_k f_{n_k},$ where $P_k$ is the natural projection of $\X$ onto $\X_k$. \endroster In particular, $\{ g_k \}$ is also scalarly null in measure. \endproclaim \demo{Proof of Lemma 3.7} Let $\{ x_n \}_{n\geq 1}$ be a normalized basis for $\X$ and let $\{ x^*_n \}$ be the corresponding biorthogonal functionals. Consider a sequence $\{ f_n \}_{n\geq 1}$ of $L_0(\X)$ functions that is scalarly null in measure. It suffices to construct inductively two increasing sequences $\{ n_k \}_{k\ge1}$ and $\{ m_k \}_{k\ge0}$ of integers and a sequence $\{ g_k \}$ of functions such that, for $\X_k \equiv \span\{x_i: m_{k-1} < i \le m_k \} $, each $g_k$ satisfies (iii) and $$\mu(\{\omega: \|g_k(\omega)-f_{n_k}(\omega)\|\ge 2^{-k}\})\le 2^{-k}. \tag1$$ To start the induction set $m_0=0$ and $n_0 = 0$. Suppose that $k\ge1$ and that $n_i$ and $m_i$ have been chosen for $i\le k-1$. Since $\{ f_n \}$ is assumed to be scalarly null in measure, it follows that, for each fixed $i$, the sequence $\{ x_i^*(f_n) \}_n$ converges to zero in measure. So there exists $n_k > n_{k-1}$ such that $$ \mu\left( \left\{ \omega: \sum_{i=1}^{m_{k-1}}|x_i^*(f_{n_k})|\ge 2^{-k-1} \right\} \right) \le 2^{-k-1} \ . $$ Hence there exists $m_k > m_{k-1}$ such that, for $\X_k \equiv \span\{x_i: m_{k-1} < i \le m_k \} $, the function $g_k$ as given in (iii) satisfies (1), which completes the induction. \qed \enddemo \demo{Proof of Theorem 3.6} Consider a sequence $\{ f_n \}$ of $L_0(\ell_1)$ functions that is scalarly null in measure. Let $\{ e_n \}$ be the standard basis of $\ell_1$. Find a blocking $\{ \X_k\}$ of $\{ e_n \}$ and sequences $\{ g_k \}$ and $\{h_k \}$ as given by Lemma~3.7. In view of (i), it is enough to show that $\{ g_k \}$ has a subsequence that is scalarly null a\.e. With that in mind, we establish the following claim. \newline {\bf Claim.} Given $\varepsilon>0$, $$ \underset \|x^*\|\le1 \to\sup \mu\left(\left\{ |x^*(g_k)|>\varepsilon\right\}\right)\rightarrow0 $$ as $k\rightarrow\infty$. \newline {\it Proof of Claim.} Suppose not. Then there exist $\varepsilon>0$ and $x_k^* \in \ell_1^*$, with $\|x_k^*\|\le1$, such that $$ \limsup_k \mu(\{|x_k^*(g_k)|>\varepsilon\})>\varepsilon.$$ Note that $\ell_1=(\sum\oplus E_k)_1$ and so $\ell_1^*=(\sum\oplus E_k^*)_\infty$. Thus there exists $x^* \in \ell_1^*$ such that, for each $k$, the functionals $x_k^*$ and $x^*$ have identical restrictions to $\X_k^*$. Hence, $$ \limsup_k \mu(\{|x^*(g_k)|>\varepsilon\})>\varepsilon,$$ which contradicts the fact that $\{ g_k \}$ is scalarly null in measure. It follows from the claim that there exists a subsequence $\{ g_{n_k} \}$ such that $$\underset \|x^*\|\le1 \to\sup\mu(\{|x^*(g_{n_k})|>2^{-k}\})<2^{-k}.$$ Clearly $\{ g_{n_k} \}$ is scalarly null a.e. \qed \enddemo However, we know of at least one space that fails ($D_\infty$). \proclaim{Theorem 3.8} $C[0,1]$ fails property ($D_\infty$). \endproclaim \demo{Proof} Consider a sequence $\{ f_n \}$ in the unit ball of $L_\infty \left( C\left[0, 1\right]\right)$. With $\O_i = \left[0, 1\right]$, write $f_n(\cdot) \equiv f_n \left(\cdot, t\right) \: \O_1 \times \O_2 \to \Bbb R$ so that \roster \item"{(i)}" $f_n(s, \cdot ) \: \O_2 \to \Bbb R$ \quad is in $ C\left[0, 1\right]$ for almost all $s\in \O_1$ \item"{(ii)}" $\lav f_n (s,t) \rav \leqslant 1$ \quad for each $t\in\O_2$ for almost all $s\in \O_1$. \endroster Such a sequence $\{ f_n \}$ is scalarly null in measure if and only if \roster \item"{(iii)}" $f_n ( \cdot, t ) \: \O_1 \to \Bbb R$ converges in measure to the null function for each $t\in\O_2$. \endroster To see this, note that if $t\in\O_2$ is fixed, then $\left( x_t^* f_n \right) (\cdot) = f_n (\cdot, t )$ where $x^*_t \in \left( C\left[0, 1\right] \right)^*$ is the point evaluation at $t$ functional. As for the reverse implication, assume that (iii) holds and let $ x^* \equiv \nu \in \left( C\left[0, 1\right] \right)^*$ be a finite regular positive Borel measure on $\Cal B\left(\O_2\right)$. It suffices to show that $ x^* f_n(\cdot) \equiv \int_{\O_2} f_n (\cdot,t) \, d\nu(t) $ converges to the null function in $\mu$-measure. Towards this, let $\lambda = \mu \times \nu$ be the corresponding product measure on the completion $\Cal A$ of $\Cal B\left(\O_1 \times \O_2\right)$. Then~(iii) implies that $f_n (\cdot, \cdot)\: \O_1 \times \O_2 \to\Bbb R$ converges to the null function in $\lambda$-measure and hence (by~(ii)) also in $L_1\left( \O_1 \times \O_2, \Cal A, \lambda \right)$. Since $$ \int_{\O_1} \left( \int_{\O_2} \lav f_n (s,t) \rav \, d\nu(t) \right) \, d\mu(s) ~=~ \iint_{ \O_1 \times \O_2} \lav f_n (s,t) \rav \, d\lambda \rightarrow0 $$ as $n\rightarrow\infty$, it follows that the sequence $\{ l_n \}$ of $L_1$ functions given by $$ l_n(\cdot) ~\equiv~ \int_{\O_2} \lav f_n (\cdot,t) \rav \, d\nu(t) $$ converges to the null function in $\mu$-measure, which gives the result. For each positive integer $n$, let $\widetilde n$ be its binary representation as a finite sequence of $0$ and $1$'s. For $t\in \O$, let $t_3$ be its unique (nonterminating) ternary expansion into $0$, $1$, and $2$'s. For $1\leqslant k \leqslant n$, let $\Gamma(k,n)$ be the collection of all $k$-tuples $(n_1, n_2, \ldots, n_k)$ of positive integers that satisfy $1\leqslant n_1 < n_2 < \ldots < n_{k-1} < n_k = n $. For $\gamma = (n_1, n_2, \ldots, n_k)$ in $\Gamma(k,n)$, let $A_\gamma$ be the set of $t\in\O$ for which $t_3$ is of the form $$ 0. ~\widetilde n_1 ~2~ \widetilde n_2 ~2~ \widetilde n_3 ~2~ \ldots \widetilde n_{k-1} ~2~ \widetilde n_k ~ 2 ~\ldots, $$ i\.e\. $$ A_{\gamma} ~= ~(~ 0. ~\widetilde n_1 ~2~ \widetilde n_2 ~2~ \ldots \widetilde n_{k-1} ~2~ \widetilde n_k ~ 1~\overline{2} \quad,\quad 0. ~\widetilde n_1 ~2~ \widetilde n_2 ~2~ \ldots \widetilde n_{k-1} ~2~ \widetilde n_k ~ \overline{2} ~] \ . $$ For technical reasons, consider the subset $$ \widetilde A_\gamma ~= ~(~ 0. ~\widetilde n_1 ~2~ \widetilde n_2 ~2~ \ldots \widetilde n_{k-1} ~2~ \widetilde n_k ~ 20~\overline{2} \quad,\quad 0. ~\widetilde n_1 ~2~ \widetilde n_2 ~2~ \ldots \widetilde n_{k-1} ~2~ \widetilde n_k ~ 220\overline{2} ~] $$ of $A_\gamma$ along with the corresponding unions $$ A^n_k ~=~ \bigcup_{\gamma\in\Gamma(k,n)} A_\gamma \text{\qquad and \qquad} \widetilde A^n_k ~=~ \bigcup_{\gamma\in\Gamma (k,n )} \widetilde A_\gamma \ . $$ The following properties of these sets will be used: \roster \item $A^n_{k_1} \cap A^n_{k_2} ~=~ \emptyset$ \quad if $k_1 \neq k_2$; \item if $t \in \cap_j A^{n_j}_{k_j}$ for an increasing sequence $\{ n_j\}$, then $\{ k_j \}$ is also (strictly) increasing; \item if $\{ n_k \}_k$ is an increasing sequence of positive integers and \newline $t_3 = 0. ~\widetilde n_1 ~2~ \widetilde n_2 ~2~ \widetilde n_3 ~2~ \ldots $, then $t \in \widetilde A^{n_k}_k$ for each $k$. \endroster For each admissible $n$ and $k$, find a continuous function $g_k^n \: \Omega_2 \to [0,1]$ that is supported on $A^n_k$ and takes the value $1$ on $\widetilde A_k^n$. Lexicographically order the dyadic interval $\{ I_k \}_{k\geqslant 1}$ of $[0,1]$ and let $h_{k} \: \Omega_1 \to [0,1]$ be the indicator function of $I_k$. Define $f_n\: \O_1 \times \O_2 \to \Bbb R$ by $$ f_n (s,t) = \sum_{k=1}^{n} h_k(s)g^n_k (t) \ . $$ Clearly, the corresponding sequence $\{ f_n \}$ is in the unit ball of $L_\infty \left( C\left[0, 1\right]\right)$ and it satisfies conditions (i), (ii), and, by (1) and (2), also (iii). Thus $\{ f_n \}$ converges scalarly in measure to the null function. However, for any subsequence $\{ n_k \}_k$ of the positive integers, for the corresponding point $t_3 = 0 . ~\widetilde n_1 ~2~ \widetilde n_2 ~2~ \widetilde n_3 ~2~ \ldots $~, it follows from condition (3) that $f_{n_k}(s,t) = h_k (s)$, which does not go pointwise a\.e\. to the null function. \qed \enddemo We shall prove below (Corollary~4.11) that $L_1$ fails ($D_p$) for $1\leq p < \infty$, but we do not know what happens when $p=\infty$. \proclaim{Question 3.9} Does $L_1$ enjoy ($D_\infty$)? \endproclaim This question can be reformulated as a question about functions of two variables as follows. For $n\ge1$, let $f_n(s,t)$ be real-valued functions on the unit square which satisfy the following: \roster \item"(i)" $f_n(s,t)=\sum_{k=1}^{N(n)}1_{E_{n,k}}(s)g_{n,k}(t)$, where $\{E_{n,k}\}_k$ is a partition of $[0,1]$ into sets of strictly positive measure; \item"(ii)" $\int|g_{n,k}(t)|\,d\mu(t)\le1$ for all $1\le k \le N(n)$; \item"(iii)" $F_n(s)=\int_A f_n(s,t)d\mu(t) \rightarrow 0$ in measure as $n\rightarrow\infty$ for every measurable set $A\subset [0,1]$. \endroster \proclaim{Question 3.9 paraphrased} For $\{f_n(s,t)\}$ as above, does there always exist a subsequence $\{f_{n_k}(s,t)\}$ such that $\int_A f_{n_k}(s,t)d\mu(t) \rightarrow 0$ a.e. for each $A\in \Sigma$? \endproclaim \heading{4. $L_p(\X)$-Bounded Sequences}\endheading We now investigate what happens when $L_\infty(\X)$-boundedness is weakened to $L_p(\X)$-boundedness. \subhead [4.i] From scalar convergence to weak a.e. convergence \endsubhead In this subsection, we shall use Dvoretzky's theorem on the existence of almost spherical sections [Dv] to prove that for $1 \leq p <\infty$ none of the properties ($A_p$), ($B_p$) nor ($C_p$) can hold in any infinite-dimensional Banach space. Let us first recall the $q$-Pettis norm of an $L_0(\X)$ function (which might be infinite): $$ \|f\|_{\Cal P_q(\X)} = \sup_{x^*\in B(\X^*)} \left( \int_\Omega |x^* f(\omega)|^q \, d\mu\right)^{1/q} \ , $$ for $1 \leq q < \infty$. The building block used in our construction is the basic example of [DG] which we now recall (and refine slightly) for the reader's convenience. \proclaim{Proposition 4.1} Let $\X$ be an infinite-dimensional Banach space and let $E$ be a measurable subset of $\Omega$. Given $\varepsilon>0$, there exists $f\in L_\infty(\X)$ such that $\|f(\cdot)\|=1_E(\cdot)$ and $\|f\|_{\Cal P_q(\X)}<2\varepsilon^{1/q}$ for each $1\le q<\infty$. \endproclaim \demo{Proof} Since $q \to \|f\|_{\Cal P_q(\X)}$ is an increasing function for a fixed $f \in L_0 (\X)$, it suffices to consider only $2\le q<\infty$. First we prove the result for $E=\Omega$. Let $$ \{I^n_k=[(k-1)/2^n,k/2^n): n\ge0 \text{~and~} 1\le k \le 2^n\} $$ be the collection of dyadic subintervals of $\Omega$. By Dvoretzky's Theorem there exist unit vectors $\{e^n_k\}_{k=1}^{2^n} $ in $\X$ such that $$ \frac{1}{2} \left( \sum_{k=1}^{2^n} |a^n_k|^2\right)^{1/2} \le \left\|\sum_{k=1}^{2^n} a^n_ke^n_k \right\| \le 2\left(\sum_{k=1}^{2^n} |a^n_k|^2\right)^{1/2}\tag1$$ for all real numbers $a^n_k$. Define $f_n:\Omega\rightarrow\X$ by $$f_n(\omega)=\sum_{k=1}^{2^n}1_{I^n_k}(\omega)e^n_k.$$ Note that $\|f_n(\omega)\| =1$ for all $\omega\in \Omega$. Fix $x^*\in B(\X^*)$. Then (1) implies that $\left(\sum_{k=1}^{2^n} |x^*(e^n_k)|^2\right)^{1/2} \le2.$ Thus, for $q\ge2$, we have $$ \align \int_\Omega \left|x^*f_n(\omega)\right|^q\,d\mu &= \int_\Omega \left|\sum_{k=1}^{2^n} x^*(e^n_k)1_{I^n_k}(\omega)\right|^q\,d\mu \\ &= 2^{-n}\sum_{k=1}^{2^n}|x^*(e^n_k)|^q\\ &\le 2^{-n}\left(\sum_{k=1}^{2^n}|x^*(e^n_k)|^2\right)^{q/2} \le 2^{q-n}~, \endalign $$ and so $\|f_n\|_{\Cal P_q(\X)}\le 2 \cdot \left(2^{-n}\right)^{\frac{1}{q}}$, which gives the result. An analogous construction can be carried out in any set $E$ of positive measure, and the result is trivial anyhow for a set $E$ of measure zero. \qed\enddemo Now we beef up the previous result. \proclaim{Proposition 4.2} Let $\X$ be an infinite-dimensional Banach space and let $h$ be a non-negative countably-valued measurable function defined on $\Omega$. Given $\varepsilon>0$ and $1 \leq q_0<\infty$ there exists $f\in L_0(\X)$ with the following properties: \roster \item $\|f(\cdot)\|=h(\cdot)$; \item $\|f\|_{\Cal P_q(\X)} < \infty$ for each $1 \leq q<\infty$; \item $\|f\|_{\Cal P_{q_0}(\X)}<\varepsilon$. \endroster \endproclaim \demo{Proof} Write $h=\sum_{k=1}^\infty a_k 1_{E_k}$, where the $a_k$'s are positive numbers and the $E_k$'s are disjoint measurable sets. Select positive numbers $\{ \varepsilon_k\}$ such that $\sum_{k=1}^\infty a_k^q \varepsilon_k$ is finite for each $ 1 \leq q<\infty$ and $\sum_{k=1}^\infty a_k^{q_0} \varepsilon_k < \left({\varepsilon}/{2}\right)^{q_0}$. By Proposition~4\.1, for each $k$ there exists $f_k \in L_\infty(\X)$ such that $\|f_k(\cdot)\|= 1_{E_k}(\cdot)$ and $\|f_k\|_{\Cal P_q(\X)} < 2\varepsilon_k^{1/q}$. Clearly, $f= \sum_{k=1}^\infty a_k f_k$ has the required properties.\qed \enddemo \proclaim{Theorem 4.3} Let $\X$ be an infinite-dimensional Banach space and let $g$ be any non-negative measurable function that is {\it not} essentially-bounded. There exists a sequence $\{g_n\}$ in $L_0(\X)$ such that the following hold: \roster \item $$\mu(\{\omega:\|g_n(\omega)\|>t\})\le \mu(\{\omega:g(\omega)>t\})$$ for all $n\ge1$ and for all $t>0$; \item $$\sum_{n=1}^\infty \|g_n\|_{\Cal P_q(\X)}<\infty$$ for each $ 1 \leq q<\infty$; \item $\{g_n\}$ converges scalarly a\.e\. to the null function; \item for each subsequence $\{g_{n_j}\}$ there exists a set $A\subset \Omega$ of full measure such that $$\limsup_j \|g_{n_j}(\omega)\|=\infty $$ for each $\omega\in A$. \endroster In particular, no subsequence of $\{g_n\}$ converges weakly on any set of strictly positive measure. \endproclaim \demo{Proof} Let $h$ be a non-negative countably-valued measurable function on $\Omega$ which is {\it not} essentially bounded and which satisfies $h(\omega)\le g(\omega)$ for $\omega\in \Omega$. Use Proposition~4\.2 to construct a sequence $\{g_n\}$ of {\it independent} $\X$-valued random variables such that \roster \item"{(i)}" each $\| g_n \|$ has the same distribution as $h$, \item"{(ii)}" $\|g_n\|_{\Cal P_q(\X)}$ is finite for each $ 1 \leq q<\infty$, and \item"{(iii)}" $\|g_n\|_{\Cal P_{n}(\X)}\le 2^{-n}$ \ . \endroster Clearly (1) is satisfied. Condition~(2) follows from the observation that, if $N \in \Bbb N$ and $1 \leq q \leq N$, then by (ii) and (iii) $$\align \sum_{n=1}^\infty \| g_n\|_{\Cal P_q(\X)} ~&\le~ \sum_{n=1}^N \|g_n\|_{\Cal P_q(\X)} + \sum_{n=N+1}^\infty \|g_n\|_{\Cal P_n(\X)}\\ ~&\le~ \sum_{n=1}^N \|g_n\|_{\Cal P_q(\X)} + \sum_{n=N+1}^\infty 2^{-n} \quad < \quad \infty~. \endalign $$ Clearly, (3) follows from (2) using Fact~1\.5. To prove (4), fix a subsequence $\{g_{n_j}\}$. Then, for each $M>0$, $$\sum_{j=1}^\infty \mu(\{\omega:\|g_{n_j}(\omega)\|>M\}) =\sum_{j=1}^\infty \mu(\{\omega: h (\omega)>M\})=\infty$$ since $h$ does not belong to $L_\infty$. So by the Borel-Cantelli lemma $\|g_{n_j}(\omega)\|>M$ infinitely often a\.e\. \qed \enddemo An appropriate choice of the measurable function $g$ (e\.g\. $g(\omega)= |\log \omega|$) in Theorem~4.3 yields the following corollary. \proclaim{Corollary 4.4} Let $\X$ be a Banach space and let $1\le p<\infty$. The following are equivalent: \roster \item $\X$ is finite-dimensional; \item $\X$ satisfies $(A^\prime_p)$; \item $\X$ satisfies $(A_p)$; \item $\X$ satisfies $(B_p)$; \item $\X$ satisfies $(C_p)$. \endroster \endproclaim \demo{Remarks} 1. Theorem~4.3 shows that there is no analogue for scalar convergence of the {\it uniform boundedness principle}: if $\X$ is infinite-dimensional then `scalar boundedness a\.e\.' does {\it not} imply `norm-boundedness a\.e\.' \newline 2. If $\|f_n\|_{L_q(\X)}\rightarrow0$ then clearly some subsequence is $\X$-norm-null a\.e. However, condition (2) of Theorem 4\.3 suggests that searching for a non-trivial scalar integrability condition which implies weak a\.e\. convergence is probably futile. \enddemo \subhead [4.ii] From scalar convergence in measure or in $L_p$ to scalar a.e\. convergence \endsubhead In this subsection we examine the properties ($D_p$) and ($E_p$) more closely for $1 \leq p < \infty$. First we recall some notation from [Pr]. Let $\{ x_n \}$ be a basic sequence in a Banach space $\X$ with coefficient functional sequence $\{ x_n^* \}$ in $\X^*$. A family $\{ \X_n \}$ of finite-dimensional subspaces of $[x_n]$ is a {\it blocking} of $\{ x_n \}$ provided there exists an increasing sequence of integers $\{ n_k \}$ with $n_1=1$ such that $\X_k=[x_i]_{i=n_k}^{n_{k+1}-1}$ for each~$k$. For $1\le p \le\infty$, if there is a positive constant $c$ so that, for each collection of vectors $\{y_i\}_{i=1}^n$, where $y_i \in \X_i$, $$ \ln \sum_{i=1}^n y_i \rn ~\leq ~ c \ln ~ \left( \ln y_i \rn \right) ~ \rn_{\ell_p} \ , $$ resp. $$ c \ln ~ \left( \ln y_i \rn \right) ~ \rn_{\ell_q} ~\leq ~ \ln \sum_{i=1}^n y_i \rn \ , $$ then we say that the blocking $\{ \X_k\}$ satisfies an upper (resp. lower) $p$-estimate. \proclaim{Theorem 4.5} Fix $1 < q \le \infty$ and let $q^{\prime}$ be its conjugate exponent. Suppose that $\X$ has a basis $\{ x_n \}$ with the property that each blocking of this basis satisfies an upper $q$-estimate. Then $\X$ enjoys ($D_{p}$) for each $q^{\prime} \leq p \leq \infty$. \endproclaim \demo{Proof} Wlog, $p = q^{\prime}$. Fix a sequence $\{ f_n \}$ in $B(L_{p}(\X))$ that is scalarly null in measure. We need to extract an scalarly null a.e.~subsequence. To this end, let $\{ \X_k \}$, $\{ P_k \}$, $\{ g_k \}$, and $\{ h_k \}$ be as provided from Lemma~3.7. It suffices to show that $\{ g_k \}$ converges to the null function scalarly a\.e. Fix $x^* \in \X^*$ and let $x^*_k = x^* \circ P_k \in X^*$. Note that $$ \lav x^* g_k (\omega) \rav \leq \ln x^*_k \rn ~ \ln g_k(\omega) \rn_{\X} \ . $$ Thus, for $\e > 0$ fixed $$ \mu (\{ \lav x^* g_k \rav \geq \e \} ) ~\leq~ \left[ \frac{\ln x^*_k \rn}{\e} ~ \ln g_k \rn_{L_{p}} \right]^{p} \ . $$ Note that each $\ln g_k \rn_{L_{p}}$ is bounded above by $2K$ where $K$ is the basis constant of $\{ x_n \}$. Thus $$ \sum_{k=1}^\infty \mu (\{ \lav x^* g_k \rav \geq \e \} ) ~\leq \left[ \frac{2K}{\e}\right]^p \sum_{k=1}^\infty \ln x^*_k \rn^p \ . $$ Since the blocking $\{ \X_k \}$ satisfies an upper $q$-estimate (say with constant $C$), $$ \ln ~ \left( \ln x^*_k \rn \right) ~ \rn_{\ell_p} ~\leq ~ 2CK \liminf_n \ln \sum_{k=1}^n x^*_k \rn ~\leq ~2 CK^2 \ln x^* \rn \ . $$ Thus $ \sum_k \mu (\{ \lav x^* g_k \rav \geq \e \} ) < \infty$. So by Borel-Cantelli, $\{ x^* g_k \}$ converges to the null function, as needed. \qed \enddemo Minor variations in the above proof give that Theorem~4.5 remains true if the word {\it basis} is replaced by {\it finite-dimensional decomposition\/}. However, there are many spaces that fail ($E_p$) (and hence fail ($D_p$)). \proclaim{Theorem 4.6} Fix $1 < q \leq \infty$ and let $q^\prime$ be its conjugate exponent. Suppose that $\X$ contains a weakly null semi-normalized basic sequence $\{ x_n \}$ which satisfies a lower $q$-estimate. Then $\X$ fails ($E_p$) for each $1\le p < q\prime$. \endproclaim \demo{Proof} We may assume without loss of generality that $\X=[x_n]$. Fix $p\in [1,q^\prime)$ and choose $q_0 \in \left(\frac{1}{q'} , \frac{1}{p} \right)$. Let $\{ g_n \}$ be a sequence of i.i.d. random variables defined on $\Omega$ with the same distribution as $ g_0 (t) = t^{-q_0} $. Define $f_n \: \Omega \to \X$ by $$ f_n (\cdot)~ = ~g_n(\cdot)~ x_n \ . $$ Since $g_0 \in L_p$ and since $ \{ x_n \}$ is semi-normalized, $\{ f_n \}$ is an $\lpx$-bounded sequence. For each $x^* \in \X^*$, $$ \ln x^* f_n \rn_{L_p} ~=~ \lav x^* (x_n) \rav ~ \ln g_n \rn_{L_p} \ . $$ Thus $\{f_n \}$ converges scalarly in $L_p$ to the null function. Fix a subsequence $\{ f_{n_j} \}$ of $\{f_n \}$. It suffices to show that $\{ f_{n_j} \}$ is not scalarly null a\.e\. To this end, let $\{ x^*_n \}$ be the sequence of biorthogonal functionals satisfying $x^*_n (x_m) = \delta_{nm}$. Since $\{ x_n \}$ is semi-normalized and satisfies a lower $q$-estimate, it follows that $\{ x^*_n \}$ satisfies an upper $q^\prime$-estimate. Consider the element $x^*\in \X^*$ given by $x^* = \sum_j j^{-q_0}~x^*_{n_j}$, which converges in $\X^*$ since $\{ x_n^*\}$ satisfies an upper $q^\prime$-estimate. Fix $M>0$. Since $$ \mu\{ x^* f_{n_j} > M \} ~=~ \mu\{ j^{-q_0} g_{n_j} > M\} ~=~ \left[ M j^{q_0} \right]^{-\frac{1}{q_0}} ~=~ M^{-\frac{1}{q_0}} ~j^{-1} \ , $$ we see that $\sum_j \mu\{ x^* f_{n_j} > M \} = \infty$, and so by the Borel-Cantelli lemma there is a set $A$ of full measure such that if $\omega \in A$ then $|x^* f_{n_j} (\omega)| > M$ infinitely often. Thus this subsequence does not converge scalarly a\.e\. \qed \enddemo By a theorem of Prus \cite{Pr} every {\it nearly uniformly convex} space (see \cite{H} for the definition of this property) satisfies the hypothesis of Theorem~4.6 for some $1 < q < \infty$ and so we obtain the following corollary. \proclaim{Corollary 4.7} Suppose that $\X$ contains a super-reflexive or (more generally) a nearly uniformly convexifiable infinite-dimensional subspace. Then $\X$ fails ($E_p$) for some $1 < p < \infty$. \endproclaim The following two corollaries follow from the previous two theorems by considering the standard unit vector basis of $\ell_p$ and $c_0$. \proclaim{Corollary 4.8} Fix $1 < q < \infty$ and let $q^\prime$ be its conjugate exponent. Then $\ell_q$ has ($D_p$) (resp.~($E_p$)) if and only if \, $1 < q^\prime \leq p \leq \infty$. \endproclaim \proclaim{Corollary 4.9} $c_0$ has ($D_p$) (and thus ($E_p$)) for each $1 \leq p \leq \infty$. \endproclaim Note that in Corollary~4.8, if $q\downarrow1$ then $p\uparrow\infty$, which suggests that $\ell_1$ should fail $D_p$ for each $1 \leq p<\infty$. Strangely, however, the truth is the complete opposite as was proved in Theorem~3.6 above: $\ell_1$ has $D_p$ for all $1 \le p \le \infty$. Finally, we determine the range of values of $p$ for which $L_q$ satisfies $(D_p)$. \proclaim{Corollary 4.10} Fix $1 < q < \infty$ and let $q^\prime$ be its conjugate exponent. Then $L_q$ has ($D_p$) (resp.~($E_p$)) if and only if \, $ \max{(2, q^\prime)} \leq p \leq \infty$. \endproclaim \demo{Proof} Since $\ell_2 $ and $\ell_q$ each embed into $L_q$, it follows from Corollary~4.8 that $L_q$ fails ($E_p$) for $1\leq p < \max{(2, q^\prime)}$. Since $L_q$ has type $\min({2, q})$ and the Haar system $\Cal H$ forms an unconditional basis for $L_q$, each blocking of $\Cal H$ satisfies an upper $\min({2, q})$-estimate. So by Theorem~4.5, if $p = \max{(2, q^\prime)}$ then $L_q$ enjoys ($D_{p})$. \qed \enddemo Let $H_1$ denote the Hardy space of analytic functions on the unit disk in the complex plane with the usual $L_1$ norm (see e\.g\. \cite{Ru}). It is known that $H_1$ contains subspaces that are isomorphic to $\ell_q$ for $1\le q \le 2$ (see e\.g\. \cite{Di}). Hence Corollary~4.8 implies the next result (cf. Question~3.9). \proclaim{Corollary 4.11} $H_1$ (and therefore also $L_1$) fails ($E_p$) for $1 \le p < \infty$. \endproclaim \demo{Remark} We do not know of a reflexive space that satisfies ($D_1$). However, we resist making the obvious conjecture. \enddemo \head 5. Completeness \endhead In this section we prove some completeness results for the topologies of scalar convergence considered in this paper. First we recall the appropriate definitions. Let $E$ be a topological vector space. A sequence $\{x_n\}$ in $E$ is a Cauchy sequence if for every zero-neighborhood $U$ there exists $N\ge1$ such that $x_n-x_m\in U$ for all $n,m \ge N$. We shall say that $E$ is complete if every Cauchy sequence converges. \proclaim{Theorem 5.1} Let $\X$ be an infinite-dimensional Banach space. The topologies of scalar convergence in measure and scalar convergence in $L_p$ (\/$1\le p <\infty$) are incomplete. \endproclaim \demo{Proof} This result is very similar in spirit to the fact that the Pettis norm is incomplete whenever $\X$ is infinite-dimensional \cite{JK}. We refer the reader to the proof of the incompleteness of the Pettis norm that is given in \cite{DG}. The construction there, which utilizes Dvoretzky's Theorem on almost spherical sections \cite{Dv}, can easily be modified, using the estimates of Proposition~4.1, to construct a sequence of functions that is Cauchy but not convergent in the topology of scalarly convergence in measure and $L_p$ for $1 \leq p < \infty$. \qed \enddemo \proclaim{Theorem 5.2} The topology of scalar convergence in $L_\infty$ is complete if and only if $\X$ is weakly sequentially complete. \endproclaim \demo{Proof} First suppose that $\X$ is not weakly sequentially complete. Let $\{x_n\}$ be a weak Cauchy sequence that does not converge weakly and let $f_n(\omega)=x_n$ ($n\ge1$). Clearly, $\{f_n\}$ is a Cauchy sequence in the topology of scalar convergence in $L_\infty$. By Proposition~3.3, a limit of this sequence, say $f$, would have to satisfy $f(\omega)=\text{ weak-lim } f_n(\omega)$ almost everywhere. Hence $\{f_n\}$ does not converge. For the converse, suppose that $\X$ is weakly sequentially complete. Let $\{f_n\}$ be a Cauchy sequence in the topology of scalar convergence in $L_\infty$. By adapting the proof of Proposition~3.3, we see that the weak sequential completeness of $\X$ guarantees that there exists a function $f:\Omega\rightarrow \X$ such that $f(\omega)=\text{weak-lim }f_n(\omega)$ a.e. By the Pettis measurability theorem \cite{P} and Proposition~1.1, $f$ is strongly-measurable, i\.e\., $f\in L_0(\X)$. It now follows easily from the fact that $\{f_n\}$ is a Cauchy sequence that $\{f_n\}$ converges to $f$ scalarly in $L_\infty$. \qed \enddemo Of more relevance to this paper is the convergence of a pointwise-bounded or an $L_p(\X)$-bounded Cauchy sequence. We investigate this question next for the topology of scalar convergence in measure. For brevity's sake we shall say that a sequence is ``scalarly Cauchy in measure" if it is a Cauchy sequence for the topology of scalar convergence in measure. \proclaim{Theorem 5.3} Let $\X$ be a reflexive Banach space. Then each pointwise-bounded sequence $\{f_n\}$ in $L_0(\X)$ that is scalarly Cauchy in measure converges scalarly in measure. \endproclaim \demo{Proof} We may assume that $\X$ is separable and hence that $\X^*$ is separable. Arguing now as in Proposition~3.1 there exists a subsequence $\{f_{n_k}\}$ such that $\{f_{n_k}(\omega)\}$ is a weakly Cauchy sequence in $\X$ almost everywhere. Since $\X$ is reflexive it is weakly sequentially complete and so (by the Pettis Measurability Theorem and Proposition~1.1) there exists $f$ in $ L_0(\X)$ such that $f_{n_k}$ converges to $f$ weakly a\.e\., thus also scalarly in measure, which is enough. \qed \enddemo \proclaim{Theorem 5.4} Let $\X$ be a reflexive Banach space. Then each $L_1(\X)$-bounded sequence $\{f_n\}$ that is scalarly Cauchy in measure converges scalarly in measure. \endproclaim \demo{Proof} First, we may assume that $\X$ is separable. By a deep result of Zippin \cite{Z} every separable reflexive Banach space is isomorphic to a closed subspace of a reflexive Banach space with a basis. So we may assume that $\X$ is isomorphically embedded into a reflexive Banach space $\Y$ with a basis. Clearly, $\{f_n\}$ is scalarly Cauchy in measure when viewed as a sequence in $L_0(\Y)$. By Proposition~1.1, it suffices to show that $\{f_n\}$ converges to some $f$ in $ L_0(\Y)$. Since $\Y$ is reflexive, a normalized basis $\{e_k\}$ for $\Y$ is both boundedly complete and shrinking \cite{LiT}. Let $C$ be the basis constant of $\{e_k\}$. For each $n$, we can expand $f_n$ with respect to the basis $\{e_k\}$ thus: $$f_n(\omega)=\sum_k f_{n,k}(\omega)e_k.$$ For each $k$, the sequence $\{f_{n,k}\}_n$ is Cauchy in measure, and hence converges in measure to some $g_k\in L_0$. By Fatou's Lemma, we have $$\align \int_\Omega \sup_N \left\|\sum_{k=1}^N g_k(\omega)e_k\right\| \,d\mu &\le C \int_\Omega \liminf_{N\rightarrow\infty} \left\|\sum_{k=1}^N g_k(\omega)e_k\right\|\,d\mu\\ &\le C \liminf_{N\rightarrow\infty} \int_\Omega \left\|\sum_{k=1}^N g_k(\omega)e_k\right\|\,d\mu\\ &\le C \liminf_{N\rightarrow\infty} \left(\liminf_{n\rightarrow\infty}\int_\Omega \left\|\sum_{k=1}^N f_{n,k} (\omega)e_k\right\|\,d\mu\right)\\ &\le C \liminf_{N\rightarrow\infty} \left(C\liminf_{n\rightarrow\infty} \int_\Omega \left\|\sum_{k=1}^\infty f_{n,k}(\omega)e_k\right\|\,d\mu\right)\\ &\le C^2 \sup_n \|f_n\|_{L_1(\Y)}<\infty. \tag1 \endalign$$ Hence $$\sup_N\left\|\sum_{k=1}^N g_k(\omega)e_k\right\|<\infty\quad a.e.$$ Since $\{e_n\}$ is boundedly complete it follows that $f(\cdot) \equiv \sum_{k=1}^\infty g_k(\cdot)e_k$ is in $L_0( \X)$; moreover, it follows from (1) that $f \in L_1(\Y)$. Fix $y^*\in \Y^*$ and $N\ge1$. Clearly, $$y^*\left(\sum_{k=1}^N f_{n,k}e_k\right)\quad \rightarrow \quad y^*\left(\sum_{k=1}^N g_ke_k\right)\tag2$$ in measure as $n\rightarrow\infty$. Let $\alpha_n$ denote the norm of the restriction of $y^*$ to $[e_k]_{k\ge n}$. Since $\{e_n\}$ is a shrinking basis, $\alpha_n\rightarrow0$ as $n\rightarrow\infty$. Now $$\align \int \left|y^*\left(\sum_{k=N+1}^\infty f_{n,k}e_k\right)\right|\,d\mu &\le \alpha_{N+1} \int \left\|\sum_{k=N+1}^\infty f_{n,k}e_k\right\|\,d\mu\\ &\le \alpha_{N+1}(1+C)\sup_n \|f_n\|_{L_1(\Y)}\quad \rightarrow \quad 0\tag3 \endalign$$ as $N \rightarrow \infty$. Combining (2) and (3) we see that $\{f_n\}$ converges to $f$ scalarly in measure. \qed \enddemo The proof of Theorem~5.3 apparently uses only the weak sequential completeness of $\X$ and the fact that $\X^*$ has the RNP. However, by Rosenthal's $\ell_1$ theorem [Ro], these two properties are {\it equivalent} to $\X$ being reflexive. Clearly, a necessary condition for the conclusion of Theorems~5.3 and~5.4 to hold is that $\X$ is weakly sequentially complete, and when $\X$ has an unconditional basis this condition is also sufficient, as our next two results show. However, we have not been able to determine general necessary and sufficient conditions on $\X$ so that the conclusions of Theorems~5.3 and~5.4 hold. In view of the next two theorems, which establish the desired conclusions for $\ell_1$, it is clear that the method of proof of Theorem~3.2 will not be of use in this situation. \proclaim{Theorem 5.5} Let $\X$ be a weakly sequentially complete Banach space with an unconditional basis. Then each pointwise-bounded sequence $\{ f_n \}$ in $L_0(\X)$ that is scalarly Cauchy in measure converges scalarly in measure. \endproclaim \demo{Proof} Let $\{e_n\}$ be a normalized unconditional basis for $\X$. We may assume, without loss of generality, that $$\left\|\sum_n a_n e_n\right\| \le \left\|\sum_n b_n e_n\right\|\tag1$$ whenever $|a_n| \le |b_n|$ for all $n$. The fact that $\X$ is weakly sequentially complete implies that $\{e_n\}$ is boundedly complete \cite{LiT}. By assumption, $$\sup_n \|f_n(\omega)\|=M(\omega)<\infty \quad \text{a.e.} \ . \tag2$$ Also, for each $n$, we can expand $f_n$ with respect to the basis $\{e_k\}$ thus: $$f_n(\omega)=\sum_k f_{n,k}(\omega)e_k \ .$$ For each $k$, the sequence $\{f_{n,k}\}_n$ is a Cauchy sequence in the topology of convergence in measure, and hence converges in measure to some $g_k$. Now (1) and (2) imply that $$\sup_N \left\|\sum_{k=1}^N g_k(\omega)\right\|\le M(\omega) \quad \text{a.e.} \ . $$ Since $\{e_n\}$ is boundedly complete it follows that $f(\cdot)=\sum_{n=1}^\infty g_n(\cdot)e_n$ is in $L_0(\X)$. Let $h_n = f-f_n$, and so $$ h_n(\cdot)= \sum_k ( g_k-f_{n,k} ) (\cdot)~e_k \ . $$ To complete the proof of the theorem, it suffices to show that $\{h_n\}$ is scalarly null in measure. So suppose, to derive a contradiction, that $\{h_n\}$ is not scalarly null in measure. Then there exists $x^*\in S(\X^*)$ and $\varepsilon>0$ such that $$ \mu \{ \omega: |x^*h_n(\omega)| > \varepsilon\}> \varepsilon\tag3$$ for infinitely many $n$. The gliding hump argument of Lemma~3.7 yields a subsequence $\{h_{n_k}\}_k$ and a blocking $\{\X_k\}$ of the basis such that each $h_{n_k}$ satisfies~(3) and $$\mu\{\omega: \|h_{n_k}(\omega) - P_kh_{n_k}(\omega)\|> \varepsilon/4 \} <\varepsilon/4,\tag4$$ where $P_k$ is the natural projection of $\X$ onto $\X_k$. We may define $y^*\in \X^*$ by defining its action on each $x_k\in\X_k$: $$ y^*(x_k)=0\quad\text{($k$ odd)};\quad y^*(x_k)=x^*(x_k) \quad\text{($k$ even)} \ . \tag5 $$ Then by (1), we have $\|y^*\|\le\|x^*\|\le1$, and so from (3), (4) and (5) we deduce that $$ \mu \{ \omega: |y^*h_{n_k}(\omega)| > \varepsilon/4\}< \varepsilon/4 \quad\text{($k$ odd)}\tag6$$ while $$ \mu \{ \omega: |y^*h_{n_k}(\omega)| > \varepsilon/2\}> 3\varepsilon/4 \quad\text{($k$ even)}.\tag7$$ Clearly, (6) and (7) contradict the fact that $\{h_n\}$ is scalarly Cauchy in measure \qed \enddemo \proclaim{Theorem 5.6} Let $\X$ be a weakly sequentially complete Banach space with an unconditional basis. Then each $L_1(\X)$-bounded sequence $ \{ f_n \}$ that is scalarly Cauchy in measure converges scalarly in measure. \endproclaim \demo{Proof} Let $\{e_k\}$ be a normalized unconditional basis for $\X$. Let $$f_n(\omega)=\sum_k f_{n,k}(\omega)e_k$$ be the expansion of $f_n$. Now, arguing as in the first half of Theorem~5.4, it can be shown that, for each $k$, $f_{n,k}\rightarrow g_k$ in measure as $n \rightarrow \infty$ and that $f(\cdot)\equiv\sum_k g_k(\cdot) e_k$ belongs to $L_1(\X)$. Now, arguing as in second half of Theorem~5.5, one uses the unconditionality of $\{e_k\}$ to prove that $\{f_n\}$ converges to $f$ scalarly in measure. \qed \enddemo \demo{Remark} Note that Theorems 5.5 and 5.6 apply to both $\ell_1$ and $H_1$. \enddemo Finally, straightforward modifications to the proofs of Theorems~5.4 and 5.6 yield the following. \proclaim{Theorem 5.7} Fix $1\le p< \infty$. Let $\X$ be a weakly sequentially complete Banach space that is either reflexive or has an unconditional basis. 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