Differentiation of an integral (Lebesgue diffentiation theorem)

Question

I am reading Kolmogorov's textbook Introductory Real analysis, at the section 31.3. In the proof of theorem 8, he wrote "since the number of sets $E_{\alpha \beta}$ is countable, this will imply $\mu\{x:f(x)<F'(x)\}=0$". I include the page of the book here. I could not understand it why the countability implies measure zero, thought as a counterexample, I can pick two function $g_1 \equiv 1$ and $g_2 \equiv 2$, so $g_1<g_2$ and I can pick sets $E_{\alpha \beta}$ the same as in the proof, I am not about to show measure of $E_{\alpha \beta}=0$, just $\mu\{x:g_1(x)<g_2(x)\}$ can not be zero. In the proof as I understand he did not use $E_{\alpha \beta}=0$ to show $\mu\{x:f(x)<F'(x)\}=0$ but instead. Please give me a hint or explain it for me!

Answer 1

We have $$\{ x: f(x)<F'(x)\} \subset_{\alpha,\beta \in \mathbb{Q}} E_{\alpha,\beta}$$

The union is countable, so $$\mu (\{x:f(x)<F'(x)\})\le\sum_{\alpha,\beta\in \mathbb{Q}}\mu(E_{\alpha,\beta})=0$$