Formally proving that if $E[\sum_{i=1}^\infty {X_i}] < \infty$, then $\sum_{i=1}^\infty {X_i} < \infty$ almost surely Given a sequence of non-negative random variables $(X_i)_{i\in\mathbb{N}}$, I would like to show that

$$ \mathbb{E}[\sum_{i=1}^\infty {X_i}] &lt; \infty$$

implies that 

$$ \sum_{i=1}^\infty {X_i} &lt; \infty $$

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:

$\begin{align}
\mathbb{E}[\sum_{i=1}^\infty {X_i}]  =&amp; \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} &lt; \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} &lt; \infty] + \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] \end{align}$

and then say that since 

$\mathbb{E}[\sum_{i=1}^\infty {X_i}] &lt; \infty$

I can then argue that the above conditions imply that 

$\mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} &lt; \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} &lt; \infty] &lt; \infty$

$\mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] &lt; \infty$

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?
This argument is correct.  In measure theory, when you allow functions with value $\infty$, the convention is indeed that $0 \times \infty = 0$, while any positive number times $\infty$ is taken to be $\infty$.  

Thus, if a random variable is infinite on a set of measure zero, this doesn't contribute to its expected value.  But if the function takes the value $\infty$ over a set of positive measure, then its expected value is infinite.  (This is just a rephrasing of your argument.)
As it turns out the approach that I outlined above is not entirely correct. The reason why is because it has conditioned the expectation on the event $\sum_{i=1}^\infty {X_i} = \infty$, which only makes sense if $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] &gt; 0$. 

A different approach which uses the same idea but avoids this annoying technicality is to use a proof by contradiction, where we assume that 

$$\mathbb{E}[\sum_{i=1}^\infty {X_i}] &lt; \infty$$

but there exists some $\omega \in \Omega$

$$\sum_{i=1}^\infty {X_i}(\omega) = \infty$$ 

so that $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] &gt; 0$.

Accordingly, we can now condition the expectation on the value of $\sum_{i=1}^\infty {X_i}$ since $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] &gt; 0$ by assumption. 

As shown above, this argument then shows that: 

$\begin{align}
\mathbb{E}[\sum_{i=1}^\infty {X_i}]  &amp;= \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} &lt; \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} &lt; \infty] &amp;+ \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] \\
&amp;= M + \infty \\
&amp;= \infty \end{align}$

which yields a contradiction and proves the required result.