Suppose that $f: \mathbb{R} \to [0, \infty)$ is Borel measurable and satisfies $\|f\|_\infty \le 1$ and $\|f\|_1 = 1$. If $$\int_a^b \! f(x) \, dx > 0$$ for all $a < b$, does it necessarily follow that $f(x) > 0$ for almost every $x \in \mathbb{R}$? If not, does it follow that $$\int_G \! f(x) \, dx > 0$$ for $G_\delta$ sets $G$ with positive Lebesgue measure?
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The assumption $f(x)\ge 0$ is not necessary, because it follows from the Lebesgue differentiation theorem (at least a.e.). – J.R. Oct 06 '14 at 12:45
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Suppose you had a Borel set $A\subset [0,1]$ which cuts every interval on a set of positive measure, but doesn't have full measure, i.e. $\mu(A\cap I)>0$ for every interval and $\mu(A)<1$. Such a set exists, as was shown by Rudin and discussed in this answer by Nick Strehlke.
Composing rescaled translates of the characteristic function of $A$ we can find a counter-example to your questions. Let
$$f(x)=C\sum_{k\in\mathbb{Z}} 2^{-|k|} \mathbf{1}_{A}(x-k)$$
where $C$ is a constant such that $\|f\|_1=1$. The integral of $f$ on every interval is positive by construction, but $f$ is not positive almost everywhere. It also answers your second question in the negative.
