I am proving the following statement:
Given any countable collection $\{\mu_{n}\}_{n=1}^{\infty}$ of measures on a measurable space $(\Omega,\mathcal{F})$, the summation $\mu:=\sum_{n=1}^{\infty}c_{n}\mu_{n}$ is also a measure on this space, for any constants $0\leq c_{n}<\infty$.
If I am not wrong, the proof is straightforward by using Tonelli's theorem to interchange the order of the double sum.
For any $A\in\mathcal{F}$, since $\mu_{n}$ is a measure and $c_{n}\geq 0$, we have $$\mu(A)=\sum_{n=1}^{\infty}c_{n}\mu_{n}(A)\geq \sum_{n=1}^{\infty}c_{n}\mu_{n}(\varnothing)=0=\mu(\varnothing).$$ Let $\{A_{k}\}_{k=1}^{\infty}$ be a collection of disjoint sets in $\mathcal{F}$. By the countable additivity of $\mu_{n}$, we can write $$\mu\Big(\bigcup_{k=1}^{\infty}A_{k}\Big)=\sum_{n=1}^{\infty}c_{n}\mu_{n}\Big(\bigcup_{k=1}^{\infty}A_{k}\Big)=\sum_{n=1}^{\infty}c_{n}\sum_{k=1}^{\infty}\mu_{n}(A_{k})=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}c_{n}\mu_{n}(A_{k}).$$ Since $c_{n}\geq 0$, the summand $c_{n}\mu_{n}(A_{k})$ is nonnegative for each $n,k$. Hence, we can interchange the order of the above double sum: $$\mu\Big(\bigcup_{k=1}^{\infty}A_{k}\Big)=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}c_{n}\mu_{n}(A_{k})=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}c_{n}\mu_{n}(A_{k})=\sum_{k=1}^{\infty}\mu(A_{k}).$$
My question is, in the above proof, I did not use the fact that $c_{n}<\infty$ at all. To use Tonelli's theorem, we don't even need the summand to be finite, but only nonnegative. I can't either find a counterexample where $\mu$ fails to be a measure when $c_{n}=\infty$ for some $n$. Why does the statement require $c_{n}$ to be finite for all $n$?
Ok, the note I was following focused more on probability measure, so I guess he implicitly assumed that the measure it was talking about is a finite measure. So, every $\mu_{n}$ has range $[0,\infty)$, and he wants to show that $\mu$ is also a (finite) measure. This is perhaps why he just assumed $0\leq c_{n}<\infty$.
My proof works well if $c_{n}=\infty$, since the Tonelli's theorem applies even when the summand is infinite, as long as the summand is nonnegative.