How to show $F^\prime (x)=f(x)$ at every Lebesgue point of $f$?

Question

Real and Complex Analysis, Page 141 $\textbf{7.11 Theorem}$ If $f\in L^1(\Bbb R^1) $ and

$$F(x)=\int^x_{-\infty}f\,dm\qquad x\in \Bbb R$$

then $F' (x)=f(x)$ at every Lebesgue point of $f$, hence $a.e.-m $.

In the proof, it only let $\Delta x$ be elements of $\{\delta_i\}$ which is a sequence of positive numbers that converges to $0$, why is it enough? In the usual way, We have to take $0<|\Delta x|<\delta$ for some $\delta>0$.

Answer 1

Defintion $x$ is Lebesgue point if $$f(x) = \lim_{\stackrel{x\in I}{|I|\to 0}}\frac{1}{|I|} \int_{I} f(y)dm(y)$$

If $x$ is a Lebesgue point then $$ f(x) = \lim_{h\to 0}\frac{1}{|[x-h,x+h]|} \int_{[x-h,x+h]} f(y)dm(y)\\= \lim_{h\to 0^+}\frac{1}{2h} \int_{x-h}^{x+h} f(y)dm(y)\\=\lim_{h\to 0^+}\frac{1}{2h} \left(\int^{x+h}_{-\infty}f\,dm -\int^{x-h}_{-\infty}f\,dm \right)\\=\lim_{h\to 0^+}\frac{F(x+h) -F(x-h)}{2h} \\= \lim_{h\to 0^+}\frac{F(x+h)-F(x) }{2h}+\frac{F(x) -F(x-h)}{2h} \\= F'(x) $$