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Lebesgue Differentiation Theorem for $\mathbb{R}$: Let $f:[a,b]\to \mathbb{R}$ be intergable and $F(x)=\int_a^xf$. Then $F$ is differentiable almost everywhere in $[a,b]$ and $F'=f$ a.e.

Is there a (simple) proof of this result that uses basic measure theory (the Lebesgue measure on $\mathbb{R}$) and the convergence theorems for the Lebesgue integral? All the proofs I have found use the Vitalli Covering Theorem and discuss higher dimensions. That's why I think there is room for simplification

Optional
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  • Is it not satisfactory to say that $F$ in this case is absolutely continuous because $f$ is $L^1$, and then appeal to the theorem on absolutely continuous functions? I find the fact that $F$ is only differentiable almost everywhere to be a sign that a more 'elementary' proof would not be easy at all. – A Blumenthal Jan 30 '13 at 18:53
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    @ABlumenthal "appeal to the theorem on absolutely continuous functions" Is this theorem so much easier to prove? – Optional Jan 30 '13 at 19:00
  • The hard work in the one-dimensional case is the fact that a monotonic function is differentiable a.e. (this is often called the Lebesgue differentiation theorem, too). M. Botsko gave a proof of this (avoiding Vitali) in 2003 official link here, see also here. After this is established, one can do the rest of the proof elementarily, I think. The third edition of Royden's book Real Analysis has a proof in one dimension (I don't know if still is in 4th edition). – Martin Jan 30 '13 at 19:11
  • @Martin By simple I don't mean elementary. You can use the countable additivity of the lebesgue measure for example, or the dominated convergence theorem. The Vitali theorem seems to be far from that however. – Optional Jan 30 '13 at 19:16
  • I used "elementary" in this loose sense. I'm quite sure (but can't check right now) that Royden's proof can be rephrased to avoid Vitali if you substitute Botsko's argument for the first fact I mentioned. Whether it counts as simple, I don't know. – Martin Jan 30 '13 at 19:32
  • @Martin Botsko's agruments seems harder than Vitali (in terms of numbers of pages) – Optional Jan 30 '13 at 19:34
  • You asked whether one can avoid Vitali and I tried to answer that affirmatively, and I gave you a reference for the one-dimensional case. That's the best I can do. Apologies if you are unhappy with that. – Martin Jan 30 '13 at 19:37
  • I have found such a simple proof. Please see here. – Analyst Jan 03 '23 at 08:53

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