Does $\int_E f = 0$ implies $f = 0$ a.e.?

Question

I know that for nonnegative $f$ and a measurable set $E$, $$\int_E f$$ (Lebesgue integral) is identically zero iff $f = 0$ a.e. How can I show this property holds for any arbitrary $f$ (not necessarily nonnegative)? Assume $f$ is continuous everywhere.

Answer 1

You can't. For example, let $f$ be the identity map on $\mathbb{R}$ and let $E = [-1, 1]$. $\int_E fdm = 0$ but $f \not= 0$.

Answer 2

What I would do is see how you defined the f? Most books defined it as two non-negatives right $f= f^+-f^-$. Then you also said the integral is additive. Thus if you show it for non-negative you are done!