# Kolmogorov's 0-1 law - Corollary from Loève's book

Studying from the book Probability Theory by Michael Loève I came across the following corollary of Kolmogorov's 0-1 law, which is not proved:

Corollary. If $X_n$ are independent r.v.'s, then the sequence $X_n$ either converges a.s. or diverges a.s.; and similarly for the series $\sum X_n$. Moreover, the limits of the sequences $X_n$ and $(X_1 + \ldots + X_n) / b_n$ where $b_n \uparrow \infty$, are degenerate.

I don't understand the statement. I know that:

1. $\limsup X_n$ and $\liminf X_n$ are tail functions.
2. $\limsup \sum_{k=1}^n \frac{X_k}{b_n}$ and $\liminf\sum_{k=1}^n \frac{X_k}{b_n}$ are tail functions.
3. The set of elements where $\sum X_n$ converges is a tail event.
4. The set of elements where $\sum X_n$ diverges is a tail event.

Consequently, applying 0-1 law, we have:

My corollary. If $X_n$ are independent r.v.'s, then the set where $X_n$ converges and the set where $X_n$ diverges have probability 0 or 1; similarly for the series $\sum X_n$. Moreover, the if the limits of the sequences $X_n$ and $(X_1 + \ldots + X_n) / b_n$, where $b_n \uparrow \infty$, exist, then they are degenerate.

However, from the first corollary I understand that the sequences always converges or diverges a.s. It also seems that they always have limits.

• Have I understood the corollary properly?
• Is it wrong?
• Can my corollary be refined?

Their corollary said "The sequence $X_n$ either converges a.s. or diverges a.s."

Your corollary said "The set (of $\omega$) where $X_n$ converges has probability 0 or 1."

These statements are equivalent to each other, since the phrase "$X_n$ converges a.s." means that $P(\{\omega:X_n\text{ converges}\})=1$.

So, the first sentence of both corollaries say the same thing. However, the second sentence of your corollary is more accurate than the second sentence of theirs. The limits in question need not exist. If they do exist, they will exist almost surely, and the resulting random variable will be degenerate (i.e. only assume a single value almost surely).

• I have proved $P(\{\omega : X_n \text{converges}\}), P(\{\omega : X_n \text{diverges}\}) \ \in \ \{0,1\}$. In these case both probabilities could be 0 at the same time. From the book's corollary I understand that one of them is always 1, what I couldn't proof. Jan 14, 2016 at 9:29
• Thank you, now I am more confident about the second part. Jan 14, 2016 at 9:32
• The reason that the two probabilities can't be 0 at the same time is because they are for complementary events, so their probabilities must add to 1. Jan 18, 2016 at 19:45
• Could you prove that they are complementary events please? Jan 19, 2016 at 12:56
• They are complimentary by definition.. Divergent means nonconvergent Jan 19, 2016 at 15:22