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Can someone provide a hint for the proof of the fact that for the Lebesgue indefinite integral, $\int_E fd\mu(x)=0$ for all $E\in S$ where $S$ is a the $\sigma$ ring, implies $f=0$ almost everywhere? In this case $f$ is a function from $X$ into a Banach space, so the method of creating sets where $f(x) > 0$ or $f(x) < 0$ isn't possible. I considered using that approach using the norm function, but I couldn't get a useful bound on the integral to show that the measure of the sets where $f \neq 0$ must be zero.

Namely taking $\int_E fd\mu < \int_E||f||d\mu<\frac{1}{n}\mu(E)$ doesn't get me anywhere.

Related: Showing that $f = 0 $ a.e. if for any measurable set $E$, $\int_E f = 0$

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  • For $\phi \in \mathbb{B}^*$, you have $\phi(\int_E f d \mu) = \int_E \phi(f) d \mu$. Hence $\phi(f(x))$ is zero ae. – copper.hat Mar 13 '15 at 04:04
  • @copper.hat I'm not sure what that lets us do... I can see how we could use the arguments used in the real case now to prove that $\phi(f(x))$ is zero almost everywhere, but what does that tell us about the function? –  Mar 13 '15 at 04:18
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    I'm thinking... – copper.hat Mar 13 '15 at 04:40
  • This is fairly straightforward if the Banach space $B$ in which $f$ has its values has a separable dual, since then $\phi_n(f(x)) = 0$ for all $\phi$ from a dense subset of $B^*$, except on a common null set. – Hans Engler Mar 13 '15 at 04:55
  • @HansEngler I don't know what that means... I'm fairly bad at this stuff as is, and we only spent about a day on dual spaces. But any hints you could give would be very much appreciated... –  Mar 13 '15 at 04:56
  • Are there any other conditions on the Banach space? – copper.hat Mar 13 '15 at 04:58
  • No, we just have that $f\in\mathcal{L}^1(X,S,\mu,B)$. Nothing else about B. –  Mar 13 '15 at 04:59
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    Anything about $X$? (Just grasping here...) – copper.hat Mar 13 '15 at 05:03
  • Nope. Nothing about X. –  Mar 13 '15 at 05:04
  • You can assume that $B$ is separable. – Hans Engler Mar 13 '15 at 15:10
  • @HansEngler Oh I suppose so, since f is measurable? –  Mar 13 '15 at 15:30

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Following Hans' suggestion in the comments above: Since $f$ is measurable, the Pettis measurability theorem tells us that $f$ is weakly measurable and $f$ is essentially separably valued (see here).

This means there is a null set $N$ such that $Y=f(X\setminus N)$ is separable.

In particular, there are elements $y_n$ that are dense in $Y$. Then we can find unit normed $\phi_n \in \mathbb{B}^*$ such that if $y \in Y$, then $\|y\| = \sup_n |\phi_n(y)|$.

Since $\|y\| = \sup_{\|\phi\|= 1} |\phi(y)|$, for each $y_n$ we can find a unit $\phi_n$ such that $|\phi_n(y_n)| \ge (1-{1 \over n})\|y_n\|$. Now suppose $y_{n_k} \to y \in Y$. Then $|\phi_{n_k}(y)| \ge |\phi_{n_k}(y_{n_k})|-|\phi_{n_k}(y-y_{n_k})| \ge (1-{1 \over n})\|y_{n_k}\|-\|y_{n_k}-y\|$ and so $\liminf_n |\phi_{n_k}(y)| \ge \|y\|$. Hence $\sup_n |\phi_n(y)| \ge \|y\|$.

In particular, $y = 0$ iff $|\phi_n(y)| = 0$ for all $n$.

Since $\phi(\int_E f d \mu) = \int_E \phi(f) d \mu$, we see that for all $\phi \in \mathbb{B}^*$, $\phi(f(x)) = 0$ ae.

Let $N_n = (\phi_n \circ f)^{-1} (\{ 0 \}^c)$ and $A = f^{-1} (\{ 0 \}^c)$. Then I claim that $A \subset (\cup_n N_n) \cup N$.

Suppose $x \in A \setminus N$, then there is some $n$ such that $\phi_n(f(x)) \neq 0$ and so $x \in N_n$.

Finally, since each of the sets $N,N_n$ are null, it follows that $A$ is null.

copper.hat
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