Bochner integral = 0 iff $f = 0$ 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.
 A: 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.
A: 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


*

*there is $A\subseteq X$ with $\mu(X\setminus A)=0$, and $f(A)=\{f(x):x\in A\}$ is separable in $B$;

*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.
A: Here is a more elementary proof. It assumes that the range of $f$, $f[X]$, is separable (so a separable Banach space would suffice).
For any ball $B(y,r)$ in the Banach space $B$,
$$0 = \int_{\{ x:\, f(x) \in B(y,r) \}} f d\mu = \int_{\{ x:\, f(x) \in B(y,r) \}} y d\mu + \int_{\{ x:\, f(x) \in B(y,r) \}} f(x) - y\; d\mu(x).$$
The first integral on the RHS equals $y\,\mu\{ x:\, f(x) \in B(y,r) \}$ and the second integral is small, ie
$$|\int_{\{ x:\, f(x) \in B(y,r) \}} f(x) - y\; d\mu(x)| \leq \int_{\{ x:\, f(x) \in B(y,r) \}} |f(x) - y|\; d\mu(x) \leq r \,\mu\{ x:\, f(x) \in B(y,r) \}.$$
So $|y|\,\mu\{ x:\, f(x) \in B(y,r) \} \leq r\,\mu\{ x:\, f(x) \in B(y,r) \}.$ 
Now fix $r > 0$. Then for all $y$ with $|y| > r$, we must have $\mu\{ x:\, f(x) \in B(y,r) \} = 0$ by the inequality we just got. Since $\{y \in f[X]: \; |y| > r\}$ is seperable, let $\{y_1, y_2, y_3, ...\}$ be a countable dense subset of it. Then
$$\{x:\;|f(x)| > r\} \subseteq \bigcup_{i=1}^\infty \{ x:\, f(x) \in B(y_i,r) \},$$
so $\mu\{x:\;|f(x)| > r\}$ must be $0$. 
Finally, let $r$ be $1/n$ for every natural number $n$, and that will give us $\mu\{x:\;|f(x)| > 0\} = 0$.
