Let $f,g:\mathbb{R}\to\mathbb{R}$ be two Lebesgue integrable functions. If we have $$f(b)-f(a)=\int_a^bg(x)dx$$ for almost all $a,b\in \mathbb{R}$. How can we modify $f$ on a set of measure zero to make it continuous on the whole real line ?

  • $\begingroup$ You mean finding $\hat f$ continuos such that $\hat f = f$ almost everywhere? $\endgroup$
    – Ant
    Apr 5, 2016 at 22:16
  • $\begingroup$ @Ant yes, exactly. $\endgroup$
    – user165633
    Apr 5, 2016 at 22:20
  • $\begingroup$ I will type the proof tomorrow if I remember, but look at the proof that functions in the sobolev spaces $W^{1,p}$ are essentially continuos, which can probably be found on Wikipedia..The proof is exactly what you're looking for. (There are probably more "natural" places to look for it but this came up to me :D also should be easily found online ) $\endgroup$
    – Ant
    Apr 5, 2016 at 22:24

1 Answer 1


Here is an outline of how to do this.

Find $a$ such that $f(b)−f(a) = \int_a^bg(x) dx$ for almost all $b$. (You need to prove that you can do so.)

Define $F(b) = f(a) + \int_a^bg(x) dx$.

By its definition $F$ matches $f$ except on a set of measure 0. Now you have to prove that $F$ is continuous.


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