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.