Integral equals zero implies function equals $0$ a.e. Let $f$ be a nonnegative bounded measurable function on a set of finite measure $E$. Assume $\int_E f = 0$. Show that $f=0$ a.e. on $E$ So I feel like I understand why this must be true, but I want to show every single detail in perfect rigour so I truly understand. I was thinking something like going about by contradiction? By letting $f > 0$ on some set of positive measure, and therefore using monotonicity of the integral or something to find a contradiction... Will these ideas work?? Is there a better way to do this? Thanks!

Its easier to show the contrapositive statement.

If $f$ is not zero almost everywhere then $A:=\\{x\in E:f(x)>0\\}$ have positive measure. Now setting $A_n:=\\{x\in E:f(x)>1/n\\}$ its easy to check that

$$ A=\bigcup_{n\in\Bbb N}A_n\tag1 $$

and because $\\{A_n\\}_{n\in\Bbb N}$ is an increasing sequence then

$$ \mu(A)=\lim_{n\to\infty}\mu(A_n)\tag2 $$

Now because $\mu(E)$ is finite then it is also $\mu(A)$, so from $\rm (2)$ we knows that there is some $N\in\Bbb N$ such that for all $n\geqslant N$ then $\mu(A)-\mu(A_n)<\mu(A)/2$, what implies that $\mu(A_n)>\mu(A)/2$, hence

$$ \int_{E}f\geqslant \int_{A_N}f\geqslant\frac{\mu(A_N)}{N} \geqslant\frac{\mu(A)}{2N}>0 $$

$\Box$