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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.

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    possible duplicate of [showing that $g=0$ almost everywhere on $\[0,1\]$](http://math.stackexchange.com/questions/86220/showing-that-g-0-almost-everywhere-on-0-1)2012-02-04

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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.

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    I should mention that this entire argument is a different formalization of what @ncmathsadist wrote. The advantage to this is that it generalizes to higher dimensions. If you know that the integral over any ball with rational radius is zero, the function is zero by the same argument.2012-02-04
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Define $F(x) = \int_0^x f(t)\,dt.$ The function $F$ is continuous and zero at every rational. Therefore $F(x) = 0$, $0\le x \le 1.$

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    Sorry about the confusion; that is correct.2012-02-04