Given a $\sigma$-additive measurable space $(E, \Sigma, \mu)$ and a measurable function $f : E \to \mathbb{R}$.
I know that from $$\int_A f(x) \mu(dx) \geq 0 $$ for all $A \in \Sigma$ it follows $f \geq 0$ $\mu$-almost surly.
If it is known that $(E, \Sigma, \mu)$ is the Borel set on $\mathbb{R}$ with the Lebesgue measure $\lambda$, $$(E, \Sigma, \mu)=(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda),$$ and we have that
$$ \int_x^y f(z) \, dz \geq 0 $$ for each interval $[x,y]$, does this imply that $f \geq 0$ almost surly?
My first thought was that it is probably true since the set of intervals is a generating set of the Borel set. However, as mentioned in Integral condition for non-negative function this is not enough in general.
If we assume $\mu$ an outer regular Borel measure and $f\in L^1({\Bbb R},\mu)$ then the claim is correct.
Taking countable increasing unions the positivity condition also holds for open intervals. More precisely, given $I=(a,b)$, let $F_n=[a_n,b_n]$ with $a_n\rightarrow a$, $b_n\rightarrow b$. Then $1_{F_n}$ converges to $1_I$ pointwise, so using dominated convergence, $$ \int_I f \; d\mu = \lim_n \int_{F_n} f\; d\mu \geq 0$$
It is then also true for any open set (being a countable disjoint union of open intervals) in ${\Bbb R}$. By outer regularity of the measure, given $A\in \Sigma$, you may find a sequence of open sets $U_k$ containing $A$ and so that $1_{U_k}$ converges a.e. pointwise to $1_A$ a.e. The positivity condition then holds for $A$ by dominated convergence.
Alternatively one may replace the outer regularity condition by inner regularity. In that case one first shows that any compact is the countable intersection of closed sets that are finite union of intervals. Then use dominated convergence to show the inequality for compact sets. And finally, inner regularity to get the result for any measurable $A$.