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 $f>0$.
 A: Well, we assume that you're considering Riemann integral, which is equivalent to Darboux integral.
Suppose that $\int_a^bf=0$ and $f\ge0$, we'd show that $\exists x\in[a,b],f(x)=0$. Suppose $I$ is a closed interval, and $\int_I f=0$. By definition of Darboux integral, given $\epsilon>0$, there's a partition $x_0<x_1<\dotsb<x_n$ such that $\sum_k(x_k-x_{k-1})\sup_{x_{k-1}\le x\le x_k}f(x)\le\epsilon\lvert I\rvert$, therefore we can obtain some $k$ such that $\sup_{x_{k-1}\le x\le x_k}f(x)\le\epsilon$. Let $I'=[x_{k-1},x_k]$, we have $\int_{I'}f=0$ and $\sup_{x\in I'}f(x)\le\epsilon$.
Thus we can obtain a sequence of shrinking closed intervals $I_1\supseteq I_2\supseteq I_3\supseteq\dotsb$ such that $\int_{I_k}f=0$ and $\sup_{x\in I_n}f(x)\le1/n$.
Now let $x_0\in\bigcap_j I_j$, and we can conclude that $f(x_0)=0$.
Remark: If we can take advantage of some tools in measure theory, the proof should be easy. Note that $f$ is measurable. $0=\int_I f\ge\alpha\cdot m(f\ge\alpha)$ hence $m(f\ge\alpha)=0$ for each $\alpha>0$, hence $f=0$ a.e.
A: 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 $F(x)=0 =>... f=0.$
