2

Theorem: let $f:I\to \mathbb{R}$ be absolutely continuous, then $f'$ exists a.e.

As mentioned here, Rudin's Real and Complex Analysis proves the theorem using a fair bit of measure theory. The theorem, however, is very easily stated with little measure theory; one would only need said theory to define the meaning of "a.e.", which could be done as follows:

Definition: for an arbitrary $x\in\mathbb{R}$, let $P(x)$ be a statement about $x$ (e.g. "$x$ is rational", "$x$ is larger than 2", etc.). We write $$¬P:=\{x\in\mathbb{R} : P(x) \text{ is false}\}$$

We say that $P$ holds almost everywhere iff

$$\inf\left\{\sum_{k=1}^{\infty}(b_k-a_k) : ¬P\subseteq \bigcup_{k=1}^{\infty}[a_k,b_k] \right\} = 0.$$

Considering that the above is the only bit of measure theory we need to state the theorem, is it possible to prove the theorem without invoking any more measure theoretical machinery?

Sam
  • 5,208
  • @AlannRosas yes! I'll edit the question. – Sam Nov 07 '22 at 22:09
  • 2
    I don't think it is possible. Lebesgue differentiation theorem is a deep theorem. In Royden's treatment, its proof invokes Vitali covering lemma and the lemma itself is tricky. – Danny Pak-Keung Chan Nov 07 '22 at 22:17
  • Remark: If you just want $f'$ exists a.e., the condition can be weaken as $f$ is a monotone function. Under absolute continuity, $f'$ not just exists, but it is integrable and the integral recovers $f$ (i.e., Fundamental Theorem of Calculus) – Danny Pak-Keung Chan Nov 07 '22 at 22:20
  • 1
    There's a bit of a history of "elementary" arguments for this (and the version Danny Pak-Keung Chan mentions). See e.g., https://doi.org/10.1090/S0002-9939-1965-0174668-0 and https://doi.org/10.2307/3647803 (and references therein). – Jakob Streipel Nov 07 '22 at 22:24

0 Answers0