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I'll post my own answer to this unless someone beats me to it and maybe even after ten others are posted in the first ten minutes, but of course there may be many ways to prove the result, so post your own if it's different and worth seeing.

Suppose $\mu_k\ge0$ for $k=1,2,3,\ldots$ and $\displaystyle\sum_{k=1}^\infty \mu_k<\infty$.

Further suppose $X_k$, $k=1,2,3,\ldots$ are independent random variables and $X_k\sim\mathrm{Poisson}(\mu_k)$ for $k=1,2,3,\ldots$.

How does one prove $\displaystyle\sum_{k=1}^\infty X_k\sim\mathrm{Poisson}\left(\sum_{k=1}^\infty \mu_k\right)$?

(Take it to be already proved that it works for finite sums. That's been asked and answered here.)

Michael Hardy
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First argument: Let the waiting time until the next arrival in a Poisson process be exponentially distributed with expected value $1$. Partition the interval $\left(0,\mu_1+\mu_2+\mu_3+\cdots\right)$ into pairwise essentially disjoint intervals of lengths $\mu_1,\mu_2,\mu_3,\ldots$. Let $X_k$ be the number of arrivals in the interval of length $\mu_k$. ("Essentially disjoint" means we don't care about intersections of measure $0$.)

$\blacksquare{}$

Second argument: Let $\varepsilon>0$. Consider $$ \sum_{k=1}^\infty X_k = \sum_{k=1}^K X_k + \sum_{k=K+1}^\infty X_k \overset{\text{respectively}} = U+V. \tag 1 $$ Choose $K$ so large that

$$\sum_{k=K+1}^\infty \mu_k<\varepsilon.\tag 2$$

The Borel–Cantelli lemma tells us only finitely many terms of the last sum in $(1)$ are positive, so $\Pr(V=\infty)=0$, and (since it's a non-negative integer-valued random variable) Markov's inequality tells us that $\Pr(V\ne0)\le [\text{the sum in }(2)] < \varepsilon$. So

\begin{align} & \Pr\left( \sum_{k=1}^\infty X_k = x \right) = \Pr\left( \sum_{k=1}^K X_k = x\text{ or } [V\ne0\ \&\ \cdots\cdots]\right) \\[10pt] \le {} & \Pr\left( \sum_{k=1}^K X_k = x\right) +\varepsilon \end{align} where the precise content of "$\cdots\cdots$" need not trouble us much. [I just realized that the inequality $\left[\vphantom{\frac11}\Pr\left( \sum_{k=1}^K X_k = x\right) \le\text{the first expression above}\right]$ is not obvious, if indeed it's true. I've deleted it and for not there is a gap in this argument. I'll be back$\ldots$] Hence $$ \left|\frac{(\mu_1+\cdots+\mu_K)^x \exp(-(\mu_1+\cdots+\mu_K))}{x!} - \frac{(\mu_1+\cdots)^x \exp(-(\mu_1+\cdots)}{x!} \right| < \varepsilon. $$ Hence $$ \left| \Pr\left(\sum_{k=1}^\infty X_k = x\right) - \frac{(\mu_1+\cdots)^x \exp(-(\mu_1+\cdots)}{x!} \right| < \varepsilon. \tag 3 $$ Since the left-hand side of $(3)$ does not depend on $\varepsilon$ and $(3)$ is true no matter how small $\varepsilon$ gets, the expression on the left must be $0$.

$\blacksquare$

Michael Hardy
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