Proving that for $f\geq0$ on $X$, $\int_X f d\mu = 0$ iff $f = 0$ a.e. Okay, so the question is the following:
Suppose $f \geq 0$ is a measurable function on the measure space $(X,\Sigma,\mu)$. Prove that
\begin{align} \int_X f d\mu = 0 \text{ if and only if } f = 0 \text{ almost everywhere.} \end{align}
I've sort of finished the proof, but my version is not very elegant or simple. I was wondering if there is a simple proof of the statement. It's meant to be using only the basics of Lebesgue integration, i.e., simple functions, etc. Thanks!
 A: The (almost) only fact needed is
Fact:  If $0\le f\le g$, then $\int f\le \int g$.
One direction of your assertion is easy:
Suppose $f$ is nonnegative and that the set
$E=\{ x : f(x)>0\}$ has measure zero. Then we have, using the Fact: $$0\le\int f\le\int \infty\cdot\chi_E=\infty\cdot\mu(E)=0.$$

For the other direction, prove the contrapositive:
Assume the set $E=\{ x : f(x)>0\}$ has positive measure. We proceed as suggested by my comment above.  For $n$ a positive integer, define the set $E_n=\{x: f(x)>1/n \}$. Note that  
$\ \ \ 1)$ $E=\bigcup\limits_{n=1}^\infty E_n$ 
and 
$\ \ \ 2)$  $E_1\subset E_2\subset E_3\subset\cdots\,$. 
From $1)$ and $2)$ (and a result referenced by the "almost" of the first paragraph) it follows that $\mu(E)=\lim\limits_{n\rightarrow\infty} E_n$. Consequently, it follows from our assumption that $\mu(E)>0$  that there is some $n$ with $\mu(E_n)>0$.  
So, with this in hand, using the Fact again, we have:
$$
\int f \ge \int_{E_n} f\ge \textstyle{1\over n}\cdot\mu(E_n)>0.
$$


 Whether or not this is "elegant" is debatable ...
