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This problem is about integrals of functions taking values in a Banach space.

Let $f \in L^1(X,S,\mu,B)$ where $X$ is a set with a $\sigma$-algebra $S$ and a measure $\mu$. Function $f$ takes values in a Banach space $B$.

If $\int_E f(x)d\mu(x) = 0$ for all $E \in S$, prove that $f = 0$ a.e.

Note that since $f$ takes values in $B$, $\int_E f(x)d\mu(x)$ also takes values in $B$, i.e. the "$0$" is the zero of the Banach space $B$.

In case of reals, there is an ordering and considering $f^{-1}([0, \infty)) \in S$ its easy to see that $f$ must be $0$ a.e. I'm unsure how to proceed for a Banach space.

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I changed the tags, but could someone more knowledgeable than me verify they fit? –  Asaf Karagila Feb 27 '12 at 10:05
bochner integral will fit –  userNaN Feb 27 '12 at 10:07

2 Answers 2

up vote 3 down vote accepted

I assume that $B^*$ is separable. In fact it is equivalent to separability of $B$.

Consider countable dense family $F=\{\varphi_n\}_{n=1}^\infty\subset B^*$. Take arbitrary $\varphi_n\in F$, then for all $E\in S$ we have $$ \int\limits_E \varphi(f(x))d\mu(x)=\varphi \left(\int\limits_E f(x)d\mu(x)\right) =\varphi(0)=0 $$ Since $E\in S$ is arbitrary we conclude that $\varphi_n(f(x))=0$ on a set $Y_n\subset X$ such that $\mu(X\setminus Y_n)=0$. Define $Y=\bigcap\limits_{n=1}^\infty Y_n$, then $\mu(X\setminus Y)\leq \sum\limits_{n=1}^\infty \mu(X\setminus Y_n)=0$. Now consider $x\in Y$, then we proved that for all $\varphi_n \in F$ we have $\varphi(f(x))=0$. Since $F$ is dense in $B^*$, then for all $\varphi\in B^*$ we have $\varphi(f(x))=0$. Then by the corollary of Hahn-Banach theorem we obtain $f(x)=0$. Thus for all $x\in Y$ we proved that $f(x)=0$ and moreover $\mu(X\setminus Y)=0$, i.e. $f=0$ a.e.

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Isn't $\int_{(0,1]} f(x)d\mu(x) = (1,0)$ ? $(1,0)$ is not equal to $0$..(the vector $(0,0)$ I mean) –  VSJ Feb 27 '12 at 9:56
I'll edit the question to make it clearer. –  VSJ Feb 27 '12 at 9:59
Surely this is the correct idea... but the proof at present seems dodgy to me. We start with a fixed $\varphi$ and using this $\varphi$ find the set $Y$. Then in the next sentence we claim this holds for all $\varphi$. It seems to me that $Y$ could vary with $\varphi$... –  Matthew Daws Feb 27 '12 at 10:34
So I need $B^*$ to be separable, then I can find a countable dense family of functionals $\varphi_n$ and countable family of sets $Y_n$ –  userNaN Feb 27 '12 at 10:36
Yep, that would work. Ah, and now I recall - I think any Bochner integral function has to have separable range, and so a simple modification would work in general! –  Matthew Daws Feb 27 '12 at 10:41

This is an extension to Norbert's answer. I am following the book "Vector Measures" by Diestel + Uhl.

We could define a $\mu$-measurable function to be Bochner integrable if and only if $\int_X \|f\| \ d\mu <\infty$. But what does "$\mu$-measurable" mean? Well, the Pettis Measurability Theorem says that this happens if and only if

  1. there is $A\subseteq X$ with $\mu(X\setminus A)=0$, and $f(A)=\{f(x):x\in A\}$ is separable in $B$;
  2. for each $\varphi\in B^*$, the scalar-valued function $\varphi\circ f$ is measurable.

Adjusting $f$ to be $0$ on $X\setminus A$, we may assume that $f(X)$ is separable. Then we can find a countable set $(\varphi_n)$ in $X^*$ which separates the points of $f(X)$, i.e. if $y\in f(X)$ and $\varphi_n(y)=0$ for all $n$, then $y=0$. Now follow Norbert's answer. So the result is true for any $B$.

Edit: In answer to Mark: Let $B=\ell^2([0,1])$ a very much non-separable Banach space. Define $f:[0,1]\rightarrow B$ by $f(t) = e_t$, where $(e_t)_{t\in[0,1]}$ is the canonical orthonormal basis for $B$. Then for any $\varphi\in B^*$ there is a countable set $A$ such that $\varphi(e_t)=0$ if $t\not\in A$. So $\varphi(f(t))=0$ off $A$; in particular, if $[0,1]$ is given Lebesgue measure, then $\varphi\circ f=0$ almost everywhere. It follows that $f$ is Pettis integrable (see http://en.wikipedia.org/wiki/Pettis_integral ) and has zero integral over any measurable $E\subseteq [0,1]$. But of course $f$ is not zero almost everywhere.

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Considering the way (1) is used, it would be interesting to know if VSJ's proposition remains true for Pettis integrable functions. I guess it probably isn't. –  Mark Feb 27 '12 at 15:58
Hi, I removed my answer because what I originally had in mind wasn't quite right. To set up the arguments you outlined in the comments requires going through something like what you and Norbert wrote also. –  Willie Wong Feb 27 '12 at 16:07
@WillieWong: Thanks. –  Mark Feb 27 '12 at 17:47

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