Is a function satisfying these premises, ae. positive or zero? Let g be defined on $[0,1]$ such that the integral between $t_0$ and $t_1$, for all $t_0 < t_1$, is positive or zero. Does g satisfy $g(x) \ge 0$ ae. ?
If not, what if we add the continuity ?
I thought about saying that if it was negative on some measure-positive set A, then take an interval within A and obtain a contradiction. But not every measure-positive set contains an interval.

Suppose $g$ is integrable and  $\int_a^b g \ge 0$ for all $0 \le a \le b \le 1$. Then $g(x) \ge 0$ ae. $x \in [0,1]$.
Let $\phi(t) = \int_0^t g$, then the Lebesgue differentiation theorem shows that
$\phi$ is ae. differentiable with $\phi'(t) = g(t)$ ae.
Suppose $x<y$, then $\phi(y) = \phi(x)+ \int_x^y g \ge \phi(x)$, so 
$\phi$ is non decreasing, hence $\phi'(t) \ge 0$.
Hence $g(t) \ge $ ae.
