# Errata: Billingsley, Theorem 31.4

Jun. 18, 2014

This post will make sense only if you are reading along with a copy of Billingsley’s Probability and Measure, 3rd 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.