Let $f\colon (0,1) \to \mathbb{R}$ be a measurable function. Suppose there exists a $C>0$ such that for all $0<a<b<1$, we have $\int_a^b f \geq C(b-a)$. I want to show that $f \neq 0$ almost everywhere on $(0,1).$
I think I have a proof of this, but it's rather long and complicated. I'm hoping there is a much simpler proof. Any help is appreciated!
This would be trivial if $f$ was assumed to be continuous but without continuity the best way to prove this is to use Lebesgue's Theorem. This theorem says $\frac 1 {b-a} \int_a^{b} f(x)\,dx \to f(a)$ as $b \to a$ almost everywhere so we get $f \geq C$ almost everywhere.
I have assumed integrability of $f$ but I will let you explain why there is no loss of generality in assuming this.