Showing that $f = 0 $ a.e. if for any measurable set $E$, $\int_E f = 0$ Let $(X, \mathcal{B}, \mu)$ be a measure space and $f$ a measurable function on $X$ and suppose that $\forall E \in \mathcal{B}$ we have that $\int_E f = 0$.  Then I want to show that $f = 0$ almost everywhere (a.e.).


*

*Suppose for sake of contradiction that $f \ne 0$ a.e.

*Then $\not\exists E \in \mathcal{B}$ s.t. $\mu(E) = 0$ and $f(x) = 0$, $\forall x \in X - E$ 

*Then $\{x : f(x) \ne 0\} = A$ is not measure zero so that either $\mu(A) > 0$ or $A \notin \mathcal{B}$.

*Now if $\mu(A) > 0$ then it is easy to see that $\int_A f \ne 0$ so that we have a contradiction of our original hypothesis that $\forall E \in \mathcal{B}, \int_E f = 0$.

*But if on the other hand $A \notin \mathcal{B}$, I cannot no longer appeal to $\int_A f \ne 0$ since $\int_A f$ is non-sense.  So I'm having trouble with this part of the argument.
 A: The set $\{x:f(x)\neq 0\}$ is the same as $f^{-1}\big((-\infty,0)\big)\cup f^{-1}\big((0,\infty)\big)$ and therefore measurable. So the issue in 5. never occurs. 
And I think you should be more explicit in step 4.
A: Arguing by contradiction definitely works, here's the idea. 
Let $\mu(\{ x : f(x) \neq 0 \}) > 0$. 
Then we have $\{x : f(x) \neq 0\} = \{x : f(x) > 0\} \cup \{x : f(x) < 0\}$, so we must have one of these two sets have positive measure. 
Let's say its the first one (the argument for the second is analogous). 
Then $\{x : f(x) > 0\} = \bigcup \{ x : f(x) \geq \frac{1}{n}\} = \bigcup E_n$ so again one of these must have positive measure. 
So say $E_k$ has positive measure, then $f$ dominates $\frac{1}{k}$ on $E_k$ so $$\int_{E_k} f \geq \int_{E_k} \frac{1}{k} = \mu(E_k)\frac{1}{k} > 0.$$ So we have a contradiction. 
EDIT: also, to comment on your proof: as you see here we don't need to deal with whether or not $A$ is measurable, it definitely is. And also, for step $4$ of your proof $\mu(A) > 0$ does not imply that $\int_A f \neq 0$ for example if $f = 1_{(0,1)} - 1_{(-1,0)}$ 
