This post will make sense only if you are reading along with
a copy of Billingsley’s *Probability and Measure*, 3^{rd}
edition. This is for my future self and my brothers and
sisters in Billingsley who may arrive here via Google.

At the beginning of the proof of Theorem 31.4 Billingsley says

… let $A_n = \cap (A \cap I)$, where the intersection extends over the intervals $I=(u,v]$ for which $u,v \in E, 0<\lambda(I)<n^{-1}$, and

$\mu(I)<(\alpha+\epsilon)\lambda(I)$

A little thought shows that this cannot possible work. Take $F(x)=x$, $\alpha=2$, $A = \mathbb{R}$ and $E=\mathbb{Q}$. Then the intersection defined in $A_n$ will be empty for all $n$ since for any $x\in \mathbb{R}$ there is an arbitrarily small interval $I_x$ with rational end-points such that $x \notin I_x$.

I think that the proof can be fixed if we replace the definition of $A_n$ by

… let $A_n = \cap (A \cap I^c)$, where the intersection extends over the intervals $I=(u,v]$ for which $u,v \in E, 0<\lambda(I)<n^{-1}$, and

$\mu(I)\ge(\alpha+\epsilon)\lambda(I)$

with the understanding that if there is no $I$ meeting these conditions then $A_n=A$.

The rest of the proof of part (i) can continue as it is. Under our new definition $A_n$ is still a Borel set and it is now actually the case that $A_n \uparrow A$.

After equation (31.20) we can argue that if $I_{nk}$ does meets $A_n$ it cannot be one of those $I$ over whose complements the intersection is taken in the definition of $A_n$. But since $I_{nk}$ has end-points in $E$ and $\lambda(I_{nk})<n^{-1}$ it must then necessarily be the case that

$\mu(I)<(\alpha+\epsilon)\lambda(I)$

This is what we need for the next displayed equation in the book.

The proof of part (ii) can be modified similarly.