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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, $\sigma\left(\bigcap_{n\ge 1}\bigcup_{m\ge n} X_m\right)\subset\bigcap_{n\ge 1}\sigma\left(\bigcup_{m\ge n}\mathcal\sigma(X_m)\right)$ ?

Thanks in advance.

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

1 Answer 1

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We use the following two facts:

  • 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.
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  • $\begingroup$ what do you mean with $\sigma(Y_n,n\ge 1)$ ? $\endgroup$
    – inequal
    Aug 12, 2014 at 15:21
  • $\begingroup$ The smallest $\sigma$-algebra making all the random variables $Y_1,\dots, Y_n,\dots,$ measurable. $\endgroup$ Aug 12, 2014 at 16:21

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