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!!!