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.).

  • $\begingroup$ 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? $\endgroup$
    – Azur
    Mar 7, 2022 at 10:19
  • 1
    $\begingroup$ @Azur You will require approximation of Borel sets by open sets. $\endgroup$ Mar 7, 2022 at 10:21
  • $\begingroup$ Yeah I actually I just realized we do need something like that to go from open to measurable sets. Cheers! $\endgroup$
    – Azur
    Mar 7, 2022 at 10:22
  • $\begingroup$ Reducing it to the case $f \geq 0$ is impossible. $\endgroup$ Mar 7, 2022 at 10:25
  • $\begingroup$ @KaviRamaMurthy Why would such a reduction be impossible? $\endgroup$
    – Anon
    Mar 7, 2022 at 10:31

1 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.

  • $\begingroup$ Do you mind just clarifying why your statement of: "it is clear that $\int_I f =0 $ for any interval $I$..." is correct? $\endgroup$
    – Anon
    Mar 7, 2022 at 12:56
  • $\begingroup$ @Anon Here we just use the property $\int_x^y=\int_a^y-\int_a^x$. $\endgroup$ Mar 7, 2022 at 16:31

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