Without Lebesgue Everyone knows following problem.
Let $f$ be positive function on $[0,1]$ and there exist $I = \int_{0}^{1}f(x)dx$. Prove that $I>0$. (recall that there are only two cases: $I=0$ or $I>0$. NOT $I<0$!!!!)
My question is how to prove it without word "Lebesgue"?? I'm interested in "simple" proof. 
Original proof.
let $I=0$. Let $A_{n} = \{x: f(x) > \frac{1}{n}\}$. It is clear that Lebesgue measure of $A_{n}$ is 0. Let $A=\{x: f(x) > 0\}$. Than $A=A_{1} \cup A_{2} \cup ...$ and of course measure of A is 0. That means that f almost everywhere equals to 0. Contradiction with $f>0$.

since f is positive . the function $F(x)=\int_{0}^{x}f(t)dt$ is an increasing funcion . and so since $0\leq F(x)\leq \int_{0}^{1}f(t)dt.$
If $I=0 $ then necesarly $F(x)=0 =>... f=0.$
