Derivative of Lebesgue integral function at the endpoints

Question

Let $f$ be a non-decreasing Lebesgue-integrable real function on $[a,b]$. I read in Kolmogorov-Fomin's Элементы теории функций и функционального анализа (p. 340 here) that $$\lim_{h\to 0^+}\Bigg(\frac{1}{h}\int_{[b,b+h]}fd\mu-\frac{1}{h}\int_{[a,a+h]}fd\mu\Bigg)=f(b)-f(a^+)$$where I have written $f(a^+)$ for $\lim_{h\to 0^+}f(a+h)$.

I know that for any summable function $f:[a,b]\to\mathbb{R}$ the equality $\frac{d}{dx}\int_{[a,x]}fd\mu=f(x)$ holds for almost all $x\in[a,b]$, but I do not see how it necessarily holds, at least in this particular case where $f$ is non-decreasing, for $x=a$ and $x=b$. Could anybody explain the reason why, in the case where $f$ is non-decreasing, $\lim_{h\to 0^+}(\frac{1}{h}\int_{[b,b+h]}fd\mu-\frac{1}{h}\int_{[a,a+h]}fd\mu)=f(b)-f(a^+)$ holds? I thank you all!!!

Answer 1

The first integral extends outside of the original domain of $f$. Presumably they are defining $f(x) = f(b)$ for all $x > b$. This implies $$ \frac 1h \int_{[b,b+h]} f \, d\mu = f(b)$$ for all $h > 0$.

On the other hand, $\lim_{x \to a^+} f(x)$ exists since $f$ is nondecreasing and equals, by definition, $f(a^+)$. Given $\epsilon > 0$ there exists $\delta > 0$ so that $a < x < a+ \delta$ implies that $|f(x) - f(a^+)| < \epsilon$. In particular, $$ 0 < h < \delta \implies \left| \frac 1h \int_{[a,a+h]} f(x) \, d\mu - f(a^+) \right| < \epsilon$$ which quantifies the limit you are looking for.