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

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.).

• Is there no easier way to do it @KaviRamaMurthy? Any open set in $\mathbb{R}$ can be written as a countable union of open intervals. And since $\int f = 0$ on any such interval, it is $0$ on any open set in $[a,b]$, from which we can prove that $f = 0$ a.e, can't we?
– Azur
Mar 7, 2022 at 10:19
• @Azur You will require approximation of Borel sets by open sets. Mar 7, 2022 at 10:21
• Yeah I actually I just realized we do need something like that to go from open to measurable sets. Cheers!
– Azur
Mar 7, 2022 at 10:22
• Reducing it to the case $f \geq 0$ is impossible. Mar 7, 2022 at 10:25
• @KaviRamaMurthy Why would such a reduction be impossible?
– Anon
Mar 7, 2022 at 10:31

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.
• Do you mind just clarifying why your statement of: "it is clear that $\int_I f =0$ for any interval $I$..." is correct?
• @Anon Here we just use the property $\int_x^y=\int_a^y-\int_a^x$. Mar 7, 2022 at 16:31