# A wrong corollary of Kolmogorov's $0-1$ Law

I either broke Probability Theory, or what is more likely, I am confused.

Let $$X_{n}$$ be a sequence of i.i.d random variables.

Then by Kolmogorov's $$0-1$$ Law, the probability that $$S_{n}=\sum_{k=1}^n X_{k}$$ converges is $$\in \{0,1\}$$.

Furthermore, the random variables $$\lim \sup S_{n}$$ and $$\lim \inf S_{n}$$ are measurable with respect to the terminal sigma-field generated by the $$X_{n}$$ and hence a.s. constant (a consequence of the 0-1 law) - is this true?

Now if $$S_{n}$$ converges a.s. then $$\lim \sup S_{n} = \lim \inf S_{n}$$ a.s.

Does this mean that if $$S_{n}$$ converges it does so to a constant?

It can't...

• Surely $\limsup S_n$ can't be measurable with respect to the terminal sigma - field , because each $S_n$ depends on $X_1$, so for example $\limsup S_n$ can't be a Borel function of just $X_2,X_3,....$, hence is not even $\sigma(X_2,X_3,...)$ measurable. Sep 12, 2020 at 15:49

By the Hewitt–Savage 0-1 law, $$\limsup S_n$$ is a constant $$C\in[-\infty,\infty]$$ ($$\because \limsup S_n$$ is measurable w.r.t. the exchangeable $$\sigma$$-algebra). Since $$S_n\overset{d}{=} S_{n+1}-X_1$$, $$C=C-X_1$$ a.s. There are two possibilities for the last equality to hold: (1) $$|C|<\infty$$ and $$X_1=0$$ a.s., (2) $$C=\pm\infty$$. The same is true for $$\liminf S_n$$.

If $$S_n$$ converges a.s. then $$\limsup S_n = \liminf S_n$$ a.s.

So far, so good.

Does this mean that if $$S_n$$ converges it does so to a constant?

That the answer is "yes" seems surprising, until you remember that a conditional statement "If $$P$$, then $$Q$$" is vacuously true when $$P$$ is false.

In your example, $$S_n$$ does not converge unless each of the $$X_n = 0$$ a.s., in which case, $$S_n = 0$$ a.s. for all $$n$$ as well and therefore we trivially have $$S_n \to 0$$ a.s.

If $$P(X_n = 0) \neq 1$$, then since $$S_n$$ does not converge, the "if" part of "if $$S_n$$ converges it does so to a constant" is false, and so the conditional statement is vacuously true.

• "$S_n$ does not converge unless each of the $X_n = 0$ a.s." does seem incorrect. It could be that $S_n$ converges by coincidence because each of the $X_n = 0$ due to some extremely unlikely form of coincidence. However the slightly modified statement "$S_n$ does not converge a.s. unless $X_n = 0$ a.s." is true of course. I think this distinction is at the heart of the OPs confusion although only the OP self can confirm/deny this.. Sep 13, 2020 at 11:10
• @Vincent I think it's clear, from my answer and in the context of OP's question, that what I meant by "$S_n$ does not converge unless..." was really "$S_n$ converges with probability zero/fails to converge with probability one unless..." Ironically, I phrased it the way I did because I was trying to avoid verbal confusion when connecting the example with vacuous truth of conditional statements. If it ain't one thing, it's another, lol. Sep 13, 2020 at 19:30

The abbreviation 'a.s' stands for 'almost surely', right? I think the correct statement is not 'if $$S_n$$ converges it converges to a constant' but 'if $$S_n$$ converges with probability 1 then it converges to a constant'.

If the probability of convergence is zero then you may happen to witness such a probability zero event and this limit can be anywhere, but there is not much reason to worry about this scenario because almost surely it won't happen.

On the other hand the situation where the probability of convergence is 1 is quite different and there we don't have just probability 1 of convergence but probability one of convergence to a limit that can be known beforehand for pretty much the reason you describe.

Unless I'm confused as well...