Does the following fact have a name?

Theorem 1 Suppose ${X_1,\ldots,X_n,\ldots}$ are a countable collection of 0/1 random variables over a probability space ${(\Omega, {\cal B}, \mathop{\mathbb P})}$ such that for every integer ${n}$ and bits ${b_1,\ldots,b_n}$ the event

$\displaystyle X_1 = b_1 \wedge \cdots \wedge X_n = b_n$

is measurable.

Suppose that

$\displaystyle \sum_{i=1}^\infty \mathop{\mathbb E} X_i \ \ \ {\rm converges}$

Then

$\displaystyle \mathop{\mathbb P} \left[ \sum_i X_i = \infty \right] = 0$

Proof: Intuitively, we want to say that by linearity of expectation we have ${\mathop{\mathbb E} [ \sum_{i=1}^\infty X_i ] = O(1)}$ and so by Markov’s inequality

$\displaystyle \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i = \infty\right] \leq \frac{O(1)}{\infty} = 0$

Just to make sure that this can be made rigorous, let’s belabor the proof step by step, doing everything completely from first principles.

Let ${c:= \sum_{i=1}^\infty \mathop{\mathbb E} X_i}$.

First of all, the event

$\displaystyle \sum_{i=1}^\infty X = \infty$

is measurable, because it is the countable intersection over all ${t}$ of the events ${\sum_{i=1}^\infty X \geq t}$, which are measurable because each of them is the countable union over all ${n}$ of the events ${\sum_{i=1}^n X_i \geq t}$. So suppose towards a contradiction that its probability is not zero, then there is ${\epsilon >0}$ such that

$\displaystyle \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i = \infty\right] = \epsilon$

In particular, for every ${t}$,

$\displaystyle \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i \geq t \right] \geq \epsilon$

But, by linearity of expectation, Markov’s inequality, and our assumption, we have that for every ${n}$ and every ${t}$

$\displaystyle \mathop{\mathbb P} \left[ \sum_{i=1}^n X_i \geq t \right] \leq \frac {\mathop{\mathbb E} \sum_{i=1}^n X_i}{t} = \frac{\sum_{i=1}^n \mathop{\mathbb E} X_i}{t} \leq \frac ct$

Now,

$\displaystyle \epsilon \leq \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i \geq t \right] = \lim_{n\rightarrow \infty} \mathop{\mathbb P} \left[ \sum_{i=1}^n X_i \geq t \right] \leq \frac ct$

which is a contradiction if we choose ${t > c/\epsilon}$.

Maybe we should also justify ${\mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i \geq t \right] = \lim_{n\rightarrow \infty} \mathop{\mathbb P} \left[ \sum_{i=1}^n X_i \geq t \right]}$. Define the disjoint events ${E_1,\ldots,E_n,\ldots}$ as

$\displaystyle E_n := \left( \sum_{i=1}^n X_i \geq t \right) \wedge \left( \sum_{i=1}^{n-1} X_i < t \right)$

Then

$\displaystyle \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i \geq t \right] = \mathop{\mathbb P} \left [ \bigcup_{i=1}^\infty E_i \right] = \sum_{i=1}^\infty \mathop{\mathbb P} [ E_i ]$

and

$\displaystyle \lim_{n\rightarrow \infty} \mathop{\mathbb P} \left[ \sum_{i=1}^n X_i \geq t \right] = \lim_{n\rightarrow \infty} \mathop{\mathbb P} \left[ \bigcup_{i=1}^n E_i \right]= \lim_{n\rightarrow \infty} \sum_{i=1}^n \mathop{\mathbb P}[ E_i] = \sum_{i=1}^\infty \mathop{\mathbb P}[E_i]$

$\Box$