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 $E$, $\int_E f = 0$
Best Answer
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.