$L^1$ interesting property

Question

Is it true that if $f:[0,1]\to\mathbb{R}$ is in $L^1 (0,1)$ and $$\int_0^t f(s) ds = 0, \forall \ t\in [0,1]$$, then $f(t) =0$ almost everywhere?

Answer 1

You can differentiate the integral with respect to $t$ almost everywhere, which would imply that $f = 0 $ a.e. A quick way to see this is via Lebesgue points. If $x\in (0,1) $ is a Lebesgue point of $f$, then $$ \lim\limits_{h\to 0} \frac{1}{2h}\int\limits_{x - h}^{x + h} f(t) dt = f(x). $$ The integral average equals $0$ since it is the difference of integrals over $[0,x+ h]$ and $[0, x- h]$ and both equal $0$. Hence $f(x) = 0 $ whenever $x$ is a Lebesuge point of $f$. Now, if $f\in L^1(0,1)$ almost all points $x\in(0,1)$ are Lebesgue points of $f$, hence the claim.