Doubts on showing that $g=0$ a.e. on $[0,1]$ given that $\int_p^q g(x)~d\mu =0$ for rationals $p,q\in[0,1]$ Could someone please explain to me why the this answer solves this problem 
I'd comment on the original question and answer, but I don't have enough rep to do that.  
thanks.
 A: Define $F(x) = \int_0^x f(t)\,dt.$ The function $F$ is continuous and zero at every rational.  Therefore $F(x) = 0$, $0\le x \le 1.$
A: An elementary argument based on the inner and outer regularity of the Lebesgue measure shows that every Lebesgue measurable set can be written as the union of a $G_\delta$ set and a set of measure zero.  The argument there thus shows that $\int_E gdm = 0$ for any Lebesgue measurable set E.  Now suppose that $g \neq 0$ on a set of positive measure.  It would then follow that either $g>0$ or $g<0$ on a set of positive measure. Consider the former case: if $m(\{g>0\})>0$ then since $\{g>0\}=\cup_n \{g>\frac{1}{n}\}$ by the continuity of measure it must be the case that $m(\{g>\frac{1}{n}\})>0$ for some n.  For this n, consider $0=\int_{\{g>\frac{1}{n}\}} g dm > \frac{1}{n} m(\{g>\frac{1}{n}\})$ a contradiction
By the way, there is a much cleaner proof than that one.  The Lebesgue Differentiation Theorem says that at every Lebesgue point (which is a.e.) we have $g(x)$ = $\lim_{r\to 0}\frac{1}{2r} \int_{x-r}^{x+r} g(t)dt$.  Taking limits with rational endpoints gives that $g(x)=0$ at all Lebesgue points because all of the integrals are zero.
