Why does the lebesgue differentiation theorem not work for arbitrary measures?

Question

I was looking through the proof for the Lebesgue differentiation theorem which is defined only for a lebesgue measure. If it was any other measure then it may not hold.

However, going through the proof I didn't find any places that I felt wouldn't also be true for an arbitrary measure.

Firstly, you prove the Lebesgue differentiation theorem is true for continuous functions. This seems to be true even for non lebesgue measures because we are just using the uniform continuity property.

Secondly, we used the fact that continuous functions are dense in L2 to pick a continuous function close to our arbitrary L2 function for which we want to show the general case of the theorem. Again, i don't see why the Lebesgue measure is any special.

Why would this not work for an arbitrary measure?

Answer 1

Lebesgue's differentiation theorem does hold for a much larger class of measures; it is not restricted to Lebesgue measure. The following statement is compiled from Measure Theory Vol. 1 by Bogachev (Theorem 5.8.8).

Let $\mu$ be a measure on $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))$ which is finite on all balls. If $f \in L^1(\mu)$, then $$f(x) = \lim_{r \to 0} \frac{1}{\mu(B(x,r))} \int_{B(x,r)} f(y) \, \mu(dy)$$ for $\mu$-almost every $x \in \mathbb{R}^n$.

More generally it is possible to consider measures on "nice" metric spaces, see e.g. Chapter 2 in Geometric Measure Theory by Federer for details.