How is $\mathcal F_\infty$ different from $\bigcup_{n=0}^\infty \mathcal F_n$?

Asked Aug 16 '15 at 20:45

Active Aug 16 '15 at 20:45

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Let $(X_n)_n$ be a sequence of random variables.

Define $\mathcal F_\infty := \sigma(X_0, X_1, \ldots)$ and $\mathcal F_n := \sigma(X_0, X_1, \ldots, X_n)$.

In the proof of the Kolmogorov's zero–one law, I came across $\bigcup_{n=0}^\infty \mathcal F_n$ and wondered in what relation it is to $\mathcal F_\infty$. In general, I don't expect $\bigcup_{n=0}^\infty \mathcal F_n$ to be a $\sigma$-algebra, yet I can't come up with a counterexample.

asked Aug 16 '15 at 20:45

Leo

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I think the one that is often cited is this one. – angryavian Aug 16 '15 at 20:49
For example, $X=\limsup\limits_{n\to\infty}X_n$ is $\mathcal F_\infty$-measurable and $[X\leqslant x]$ is not, in general, in $\bigcup\limits_n\mathcal F_n$. – Did Aug 16 '15 at 20:50

How is $\mathcal F_\infty$ different from $\bigcup_{n=0}^\infty \mathcal F_n$?

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