If $\sum_{n\geqslant 1}X_n$ converges almost surely then so does $\sum_{n\geqslant 1}Y_n$ Reading the book Probability and Stochastics of Erhan Çinlar in page 126 he states (paraphrasing a bit) the following:

Let $(X_n)$ a bounded sequence of independent random variables and suppose that $\sum_{n\geqslant 1}X_n$ converges almost surely. Now suppose that $(Y_n)$ is independent of $(X_n)$ and with the same distribution, then $\sum_{n\geqslant 1}Y_n$ also converges almost surely.

This seems to me a completely erroneous assertion, there is no reason to think that $\sum_{n\geqslant 1}Y_n$ converges almost sure, otherwise by Skorokhod theorem every sequence of random variables that converges in distribution will also converges almost surely, what is clearly false.
I'm right or there is something that I'm not seeing here?

UPDATE: ok, I didn't read carefully the text as the statement seems true as far the set defined by $B:=\{x\in \mathbb{R}^{\mathbb{N}}:\sum_{n\geqslant 1}x_n\in \mathbb{R}\}$ is measurable in the product $\sigma $-algebra of $\mathbb{R}^{\mathbb{N}}$. Then seeing the sequences $(X_n)$ and $(Y_n)$ as random objects $X,Y:\Omega \to \mathbb{R}^{\mathbb{N}}$ then if both have the same law then clearly we will have that $\Pr [X\in B]=\Pr [Y\in B]=1$.
 A: There's a theorem by Levy . See thm 7.3.2 in Sidney Resnick: A Probability Path which says that for independent $X_{n}$'s , $S_{n}$ converges almost surely iff $S_{n}$ is almost surely Cauchy iff $S_{n}$ converges in Probability iff $S_{n}$ is  Cauchy in Probability.  The reason I am posting it as an answer because it is long enough and that it does not depend on the boundedness of $X_{n}$'s which allows you to conclude using Kolmogorov's Three Series Theorem.
Here by Cauchy in probability , I mean Cauchy in measure. i.e. for every $\epsilon>0$ $P(|S_{n}-S_{m}|>\epsilon)\xrightarrow{m,n\, \to\infty} 0$. i.e. for $\epsilon,\delta >0$ . There exists $N$ such that $m,n\geq N\implies P(|S_{n}-S_{m}|>\epsilon)<\delta$
Let $S_{n}^{X}=\sum_{i=1}^{n}X_{i}$ and $S_{n}^{Y}=\sum_{i=1}^{n}Y_{i}$ .
Then $S_{n}$ converges almost surely $\iff$ $S_{n}$ is Cauchy in Probability . So for $\epsilon,\delta>0$ , there exists $N$ such that $$P(|S^{X}_{n}-S^{X}_{m}|>\epsilon)<\delta\\ \iff P(|X_{m+1}+...+X_{n}|>\epsilon)<\delta \iff \\P(|Y_{m+1}+...+Y_{n}|>\epsilon)<\delta\iff P(|S_{n}^{Y}-S_{m}^{Y}|>\epsilon)<\delta$$.
Thus $S_{n}^{X}$ is Cauchy in Probability if and only if $S_{n}^{Y}$ is Cauchy in Probability. And thus by Levy's Theorem $S_{n}^{X}$ converges a.s. if and only if $S_{n}^{Y}$ converges a.s.
