# Why is $\lim_{n\rightarrow\infty}\frac{1}{n!}\sum_{\rho\in S(n)}\varphi\left(X^\rho\right)$ measurable w.r.t. the tail $\sigma$-algebra?

Let $X=\left(X_1, X_2, \dots\right)$ be an exchangeable family of random variables with values in a Polish space $E$. Fix $k\in\mathbb{N}$ and let $\varphi:E^k\rightarrow\mathbb{R}$ be measurable with $E\left(\left|\varphi\left(X_1,X_2, \dots, X_k\right)\right|\right)<\infty$. Without risk of ambiguity, denote $\varphi(X):=\varphi\left(X_1,X_2,\dots,X_k\right)$ and let $A_n\left(\varphi\right):=\frac{1}{n!}\sum_{\rho\in S(n)}\varphi\left(X^\rho\right)$ ($S(n)$ being the set of permutations of the first $n$ natural numbers and $X^\rho:=\left(X_{\rho(1)},X_{\rho(2)},\dots\right)$).

Denoting by $\mathcal{T}$ the tail $\sigma$-algebra of $X$, [Klenke] claims (Theorem 12.17, p. 238) that $\lim_{n\rightarrow\infty}A_n\left(\varphi\right)$ is $\mathcal{T}$-measurable. Here's the argument given in the book.

It can be easily verified that

$$\lim_{n\rightarrow\infty}\frac{\#\left\{\rho\in S(n) : \rho^{-1}(i)\leq l\space\mathrm{for}\space\mathrm{some}\space i\in\left\{1,\dots, k\right\}\right\}}{n!}=0\space\space\mathrm{for}\space\mathrm{all}\space l\in\mathbb{N}$$

Thus, for lagre $n$, the dependence of $A_n\left(\varphi\right)$ on the first $l$ coordinates is negligible. $\square$

I understand why this argument works if $\varphi$ is bounded, but not the general case. Any help will be appreciated.

References Klenke, Achim. "Probability Theory", 2008

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A general proof technique in analysis, is if you know something is true for all bounded functions, you can extend it to all integrable functions, by taking the limit of a sequence of bounded functions that approximate your integrable function in the limit.

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Thanks. Sounds like a good suggestion, but i don't see how to apply the idea to the present case. –  Evan Aad Mar 19 '13 at 6:12