Measure-theoretic analogue of a result from elementary calculus

Question

I recall from elementary calculus being taught about defining the "average" of a continuous function from a compact subset $K = [a, b]$ of $\mathbb{R}$ to $\mathbb{R}$ by $\frac{\int_{a}^{b} f(x) \mathrm{d}x}{b - a}$. Moreover, one can show pretty easily that if we take $b \searrow a$, then the average will go to $f(a)$. This is a pretty simple result. Note that $[a, b]$ is compact, and so $f$ is uniformly continuous on it. Thus if I pick $\epsilon > 0$, then I can find $\delta > 0$ such that if $a \leq x \leq a + \delta = b < \delta$, then $|f(x) - f(a)| < \epsilon$. Let $b = a + \delta$. Then by the mean value theorem for integrals, there must exist $c \in [a, b]$ such that $f(c) = \frac{\int_{a}^{b} f(x) \mathrm{d}x}{b - a}$, i.e. $\frac{\int_{a}^{b} f(x) \mathrm{d}x}{b - a} \in f [a, b]$, which implies $|\frac{\int_{a}^{b} f(x) \mathrm{d}x}{b - a} - f(a)| < \epsilon$. Thus taking $\delta \searrow 0$, we have that $\frac{\int_{a}^{a + \delta} f(x) \mathrm{d}x}{\delta} \to f(a)$.

However, I recall reading somewhere a similar result in the context of Lebesgue integrals. An author was dealing in an integrable function $f: [0, 1] \to \mathbb{R}$, where $[0, 1]$ was endowed with the Lebesgue probability measure $\lambda$. He claims (without proof) that for almost every $x \in [0, 1]$, we have that $\lim_{r \to 0} \frac{\int_{x}^{x + r} f(y) \mathrm{d} \lambda (y)}{r} = f(x)$. I would imagine this is a fairly standard result, probably phrased in the context of Radon measures, but I don't know how one would demonstrate it, as in the proof of the result in the sense of Riemann integration I used continuity and uniform continuity, which isn't nearly enough for what the author claims (if I recall correctly, the author made no assumptions of continuity). How would I go about showing this result, and in what generality could I state it?

Answer 1

The Lebesgue-Besicovitch Differentiation Theorem is the most general form I know of. The statement of the theorem is:

Let $\mu$ be a Radon measure on $\mathbb{R}^n$ and $f \in L^1_{loc}(\mathbb{R}^n,\mu)$. Then \begin{equation} \lim_{r \to 0}\frac{\int_{B(x,r)}fd\mu}{\mu(B(x,r))} = f(x) \end{equation} for $\mu$-a.e. $x \in \mathbb{R}^n$.

See Evans & Gariepy, Measure Theory and Fine Properties of Functions for a proof (I believe starting on p.43)