The limit $\lim_{h\to0}\frac{1}{h}\int_0^hf(s) \, ds$

Question

We know that if a function $f:\mathbb{R}\to\mathbb{R}$ is continuous then we have

$$\lim_{h\to0}\frac{1}{h}\int_0^hf(s) \, ds=f(0).$$

What can we say if $f$ is continuous almost everywhere or satisfy some weaker condition?

Answer 1

For the sake of completeness: if $f$ is locally Lebesgue integrable, then for almost every point $x$ in its domain we have $$\lim_{h\to0}\frac{1}{h}\int_x^{x+h}f(s)\,ds=f(x) \tag1$$ This is a consequence of the Lebesgue differentiation theorem, which actually has a stronger assertion:
$$\lim_{h\to0}\frac{1}{h}\int_x^{x+h}|f(s)-f(x)|\,ds=0 \tag2$$