# Limit of measurable functions from measurable spaces to complete metric spaces is measurable

Here is a lemma in the Second Edition of Foundations of Modern probability by Olav Kallenberg(page 6, lemma 1.10)

Let $f_1, f_2, \dots$ be measurable functions from a measurable space $(\Omega, \mathcal A)$ into some complete metric space $(S, \rho)$. Then $\{\omega; f_n(\omega)~\text{converges}\} \in \mathcal A$.(Of course the sigma algebra on $(S, \rho)$ is the Borel sigma algebra)

Since $(S, \rho)$ is a complete metric space, we have $$\{\omega; f_n(\omega)~\text{converges}\} = \{\omega; \lim_{n \to \infty} \sup_{m \ge n} \rho(f_m, f_n)=0 \}.$$

To show the measurability of RHS, we want to use the facts that the composition of two measurable functions is measurable and that continuous maps are measurable and that the lims and sups of measurable functions are measurable.

In this proof, I think I need to show either of the following compositions is a composition of measurable maps(the first maps are $(f_m, f_n)$ while the second maps are $\rho$): $$(\Omega \times \Omega, \mathcal A \otimes \mathcal A) \to (S\times S, \mathcal B (S) \otimes \mathcal B (S)) \to \mathbb R_+ \tag{1}$$ $$(\Omega \times \Omega, \mathcal A \otimes \mathcal A) \to (S\times S, \mathcal B (S \times S)) \to \mathbb R_+ \tag{2}$$

I think it is easy to show that the first map of (1) and the second map of (2) are measurable. How about the rest of them? Did I misunderstand the author? Is there any other proof for this lemma if such an approach doesn't work? Do we have to add the condition that $(S, \rho)$ is also separable to make the middle two terms coincide with each other?

Thanks in advance!

## 1 Answer

Actually, the first mapping is $F_{n,m}:\Omega\to S\times S$, i.e. $F_{n,m}(\omega)=(f_n(\omega),f_m(\omega))$, which is $\mathcal{B}(S)\otimes \mathcal{B}(S)$-measurable (e.g. Lemma 1.8, Kallenberg). Then $\rho \circ F_{n,m}$ is measurable if $S$ is separable (so that $\mathcal{B}(S)\otimes \mathcal{B}(S)$ coincides with $\mathcal{B}(S\times S)$); otherwise ($S$ is not separable), $\rho:S\times S\to \mathbb{R}$ need not be measurable w.r.t. the product $\sigma$-algebra.

• Thank you for pointing out my mistake! So you believe that we must add the condition that $(S,ρ)$ is also separable? – No One Dec 25 '15 at 20:45
• @TiWen It's not that obvious. Here is a counterexample (Ex. 1.86-1.87): books.google.ca/… – d.k.o. Dec 26 '15 at 5:44
• Thank you so much! But Exe.1.86&1.87 in this book are not counterexamples to the original lemma 1.10, right? I think I still need to see a proof or a counterexample to lemma 1.10 of Kallenberg's book... – No One Dec 26 '15 at 6:34
• @TiWen This example shows that $\rho$ may not be measurable w.r.t. $\mathcal{B}(S)\otimes \mathcal{B}(S)$ (for general metric space $S$). Also, $F$ is need not be $\mathcal{B}(S\times S)$-mble. – d.k.o. Dec 26 '15 at 6:51