$f \in L^1([a,b])$ and $\forall x, \int_a^x f(u)du = 0$ then $f=0$ a.e.

Question

I'm trying to prove the following claim:

If $f \in L^1([a,b])$ and $\forall x\in [a,b], \int_a^x f(u)du = 0$ then $f=0$ almost everywhere (a.e.).

I thought of using the definition of $\int f = \int f^+ - \int f^-$, with the goal of applying the following known fact for functions in $L^+$ (both $f^+$ and $f^-$ are in $L^+$):

If $f \in L^+$ and $\int f = 0$ then $f=0$ a.e.

But I cannot immediately use this since I only have that their difference is 0 (i.e. $\int f^+ - \int f^- = 0$), and not that each one of them is 0.

Any ideas on how I could proceed? (ideally using basic results in Real Analysis - e.g. convergence theorems like DCT, results on $L^+$, basic results on $L^1$ etc.).

Answer 1

Here is an attempt at an elementary proof: Consider $\mathcal F$ the family of measurable sets $E\subset[a,b]$ such that $\int_Ef=0$. From your original statement, it is clear that $\int_If=0$ for any interval $I\subset[a,b]$ so $\mathcal F$ contains the open intervals. We have that $\mathcal F$ is a Dynkin system (because the unions are only taken over disjoint families). Thus the $\sigma$-algebra generated by the open intervals--ie., the Borel sigma algebra--is in $\mathcal F$. Borel sets and Lebesgue sets differ by null sets so in fact $\mathcal F$ is all of $\mathcal L([a,b])$. Now we're essentially done; for example consider sets like $\{f>1/n\}$ which are in $\mathcal F$ and must therefore have measure zero.