Here is an application of the Kolmogorov Zero-One Law given in my textbook (a probability path by Resnick page 107-108). It states that the random variables $\limsup_nX_n$ and $\liminf_nX_n$ are constant with probability 1 due to the law mentioned before. What I do not understand here is that the law only states that these two events should have probability 0 or 1. Could anyone clarify this, please? Thank you!


For a fixed $\lambda$, define the event $A:=\{\limsup_{n\to \infty}X_n\leqslant \lambda\}$. This is a tail event, hence it has probability $0$ or $1$. If we define $Y:=\limsup_{n\to \infty}X_n$, we thus have that for each $\lambda$, $\mathbb P\{Y\leqslant \lambda\}\in \{0,1\}$. Defining $c:=\inf\{\lambda, \mathbb P\{Y\leqslant \lambda\}=1\}$, we get that $Y=c$ almost surely.

The reasoning is similar for the liminf.

  • $\begingroup$ My question is why it must be 1 in this case. @DavideGiraudo $\endgroup$ – mogs Sep 28 '14 at 10:20
  • $\begingroup$ Well, don't you have $\mathbb P(Y=c)=1$? $\endgroup$ – Davide Giraudo Sep 28 '14 at 10:22
  • $\begingroup$ super thanks @DavideGiraudo :D $\endgroup$ – mogs Sep 28 '14 at 10:25
  • $\begingroup$ last question, @DavideGiraudo, how can you show that event A is a taik event... :-) $\endgroup$ – mogs Sep 28 '14 at 10:31
  • $\begingroup$ Notice that $\limsup_n X_n=\limsup_n X_{n+j}$ for each integer $j$, and the random variable in the right is measurable with respect to $\sigma(X_k,k\geq j)$. $\endgroup$ – Davide Giraudo Sep 28 '14 at 10:49

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