# Why is $\limsup\limits_{n\to\infty}X_n$, $C_{\infty}$-measurable?

Why is $$\limsup\limits_{n\to\infty}X_n$$, $$C_{\infty}$$-measurable ?

If $$\mathcal B_n=\sigma(X_n)$$,$$\quad\mathcal C_n=\sigma\left(\bigcup_{m\ge n}\mathcal B_n\right)$$,$$\quad\mathcal C_\infty=\bigcap_{n\ge 1}\mathcal C_n$$ and $$X_n's$$ are independent.

According to the definiton above $$\mathcal B_n$$ is the smallest $$\sigma$$-algebra, which makes $$X_n$$ measurable and if I write

$$\limsup\limits_n X_n=\bigcap_{n\ge 1}\bigcup_{m\ge n} X_m$$

so do I have to show that, $$\sigma\left(\bigcap_{n\ge 1}\bigcup_{m\ge n} X_m\right)\subset\bigcap_{n\ge 1}\sigma\left(\bigcup_{m\ge n}\mathcal\sigma(X_m)\right)$$ ?

• After the "so do I have to show that...", what is the meaning of the union of the $X_m$'s? Aug 12, 2014 at 15:09
• @Davide Giraudo, on the LHS is the smallest sigma-algebra which makes $\limsup X_n$ measurable and on the right $\mathcal C_{\infty}$ Aug 12, 2014 at 15:11
• The question is to understand the meaning of the (quite unorthodox) identity $$\limsup\limits_n X_n=\bigcap_{n\ge 1}\bigcup_{m\ge n} X_m.$$
– Did
Aug 12, 2014 at 16:03

• for a (deterministic) sequence of real numbers $(a_n)_{n\geqslant 1}$, we have $\limsup_na_n=\limsup_na_{n+k}$ for any integer $k$;
• if $(Y_n)_{n\geqslant 1}$ is a sequence of random variables, then $\limsup_n Y_n$ is $\sigma(Y_n,n\geqslant 1)$ measurable.
• what do you mean with $\sigma(Y_n,n\ge 1)$ ? Aug 12, 2014 at 15:21
• The smallest $\sigma$-algebra making all the random variables $Y_1,\dots, Y_n,\dots,$ measurable. Aug 12, 2014 at 16:21