Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space. Let $X,Y$ random variables. We suppose $\Omega $ discrete. We write $$X=\sum_{i=1}^\infty a_i\boldsymbol 1_{\{X=a_i\}}$$ and $$Y=\sum_{k=1}^\infty b_k \boldsymbol 1_{\{Y=b_k\}}.$$
We want to write $X+Y$ as a sum of $\sum_{k}x_k\boldsymbol 1_{A_k}$. At a moment in the proof, they do :
$$\sum_{k=1}^\infty a_k\boldsymbol 1_{\{X=a_k\}}\sum_{j=1}^\infty \boldsymbol 1_{\{Y=b_j\}}=\sum_{k=1}^\infty \sum_{j=1}^\infty a_kb_j\boldsymbol 1_{\{X=a_k,Y=b_j\}}=\sum_{j=1}^\infty \sum_{k=1}^\infty a_kb_j\boldsymbol 1_{\{X=a_k,Y=b_j\}}.$$ In the last equality they permute the two sum. Don't we need condition to do that ? Like $\sum_{i,j}a_ib_j$ converge uniformly ? Because nothing has been mentioned at this subject.
For a fix $\omega \in \Omega $, $$\sum_{j=1}^\infty \sum_{i=1}^\infty a_ib_j\boldsymbol 1_{\{X=a_i, Y=b_j\}}(\omega ),$$ is a finite sum.