I am quite confused with this problem. This is somehow similar to a proposition following the Chebychev's Inequality. However, the latter doesn't require to $f$ to be bounded.
Do I just use similar approach but intead of having $E_n=\{x\in E:f(x)>=1/n\}$, I would have $E_n=\{x\in E: f(x)<=1/n\}$? Help, I am confused with the boundedness of f.
Should I use the Simple Approximation Lemma that assumes $f$ to be bounded? How do I go about with this?
On
The missing hypothesis in your statement is that $f$ must also be assumed to be nonnegative.
In this case, it is the contrapositive of the following statement:
If $f$ is a bounded measurable function such that $f>0$ on a set $E$ of positive measure, then $\int_E f>0$.
To prove this, let $A_n=\{x:f(x)\ge 1/n\}$. If there is some $A_n$ with positive measure, then $$\int_E f\ge\int_{A_n} f\ge\int_{A_n}1/n=\frac{\mu(A_n)}n>0,$$ otherwise the set $E=\{x:f(x)>0\}=\bigcup_{n\in\Bbb N}A_n$ is a countable union of nullsets and hence null, contradicting the assumption.
Since $f# is bounded then I could use the definition of the integral of bounded functions, that is, $\int_{E}{f} = inf{\int_{E} \psi : \psi simple, f<=\psi on E} $ which is equal to 0 since $\int_{E}{f}$ is 0. Then $\int_{E}{\psi} = 0$ Thus, $\psi = 0$ and therefore $f=0$ almost everywhere on E.
But I am not convinced with my own work...