# Probability distribution and convergence almost surely

I'm trying to understand the Kolmogorov's three series theorem. Studying the proof I bump into the following proposition:

Let $\{X_n\}_{n\in \mathbb{N}}, \{Y_n\}_{n\in \mathbb{N}}$ be two sequences of real independent r.v. defined on the same probability space (the variables $\{Y_n\}$ are also independent from the $\{X_n\}$) and uniformly bounded (i.e. $\exists \lambda>0 \ s.t. |X_n(\omega)|\leq \lambda$ for all $\omega \in \Omega, n\in \mathbb{N}$). Suppose that $Y_n$ has the same probability distribution of $X_n$, and that $\sum_{n=1}^{+\infty} X_n$ convergers almost surely, then also $\sum_{n=1}^{+\infty} Y_n$ converges almost surely.

How can I prove this?

I understand that since $X_n$ and $Y_n$ have the same distribution and $\sum_{n=1}^{+\infty} X_n$ converges a.s. then it must converge also in probability and then $\sum_{n=1}^{+\infty} Y_n$ must converge in probability but this does not imply, in general, that $\sum_{n=1}^{+\infty} Y_n$ converges a.s.

Thanks for the help.

• It may be worth noting a theorem of P. Lévy: for a series of independent random variables, convergence in probability implies almost sure convergence. – John Dawkins Sep 13 '16 at 17:57

Note that the event $\left\{\omega\in\Omega, \sum_{l=1}^{+\infty}X_l\right\}$ is convergent may be written as $$\bigcap_{i\geqslant 1}\bigcup_{N\geqslant 1}\bigcap_{N\leqslant m\leqslant n}\left\{\left|\sum_{l=m}^n X_l\right|\leqslant\frac 1i \right\} .$$ Therefore, the probability of $\left\{\omega\in\Omega, \sum_{l=1}^{+\infty}X_l\right\}$ depends only on the distribution of the sequence $\left(X_l\right)_{l\geqslant 1}$.
In the context of the question, the sequences $\left(X_l\right)_{l\geqslant 1}$ and $\left(Y_l\right)_{l\geqslant 1}$ have the same distribution.