Given a sequence of non-negative random variables $(X_i)_{i\in\mathbb{N}}$, I would like to show that
$ \mathbb{E}[\sum_{i=1}^\infty {X_i}] < \infty$
implies that
$ \sum_{i=1}^\infty {X_i} < \infty $
almost surely
The approach that I have in mind is to condition the expectation on the value of $\sum_{i=1}^\infty {X_i}$ as follows:
$\begin{align} \mathbb{E}[\sum_{i=1}^\infty {X_i}] =& \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} < \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} < \infty] + \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] \end{align}$
and then say that since
$\mathbb{E}[\sum_{i=1}^\infty {X_i}] < \infty$
I can then argue that the above conditions imply that
$\mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} < \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} < \infty] < \infty$
$\mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] < \infty$
Given that $\mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] = \infty $, this must mean that $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] = 0$
Although this explanation makes sense intuitively, it doesn't seem formal enough (in particular it relies on the notion that $0 \times \infty = 0$). Is there a more elegant or formal approach?