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
$$ \mathbb{E}[\sum_{i=1}^\infty {X_i}] < \infty$$
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?
 A: 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.)
A: Let $Y$ be the sum.  You want to show that $\Pr(Y<\infty)=1$ if $E(Y)<\infty$.  If $\Pr(Y<\infty)\ne1$, then
$$E(Y) = \infty\cdot\Pr(Y=\infty)+\text{a non-negative-valued integral over some subset of the space}.$$
This is equal to $\infty$.  So what you're trying to prove is the contrapositive of that.
A: It is a truth that if $X$ is a nonnegative random variable such that $X = \infty$ with positive probability then $\int X \ dP = \infty$. Depending on how you have introduced the Lebesgue integral, this is either part of the definition (when I learned measure-theoretic probability this was the case) or an elementary fact about the integral derived at the outset. Apply this to $\sum_{i= 1} ^ \infty X_i$; if $\sum_{i = 1} ^ \infty X_i$ had positive probability of being $\infty$ then we know immediately that $E(\sum_{i = 1} ^ \infty X_i) = \infty$. 
A: 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 $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] > 0$. 
A different approach which uses the same idea but avoids this annoying technicality is to use a proof by contradiction, where we assume that 
$$\mathbb{E}[\sum_{i=1}^\infty {X_i}] < \infty$$
but there exists some $\omega \in \Omega$
$$\sum_{i=1}^\infty {X_i}(\omega) = \infty$$ 
so that $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] > 0$.
Accordingly, we can now condition the expectation on the value of $\sum_{i=1}^\infty {X_i}$ since $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] > 0$ by assumption. 
As shown above, this argument then shows that: 
$\begin{align}
\mathbb{E}[\sum_{i=1}^\infty {X_i}]  &= \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} < \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} < \infty] &+ \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] \\
&= M + \infty \\
&= \infty \end{align}$
which yields a contradiction and proves the required result. 
