Let $\{\mu_n\}_{n=1}^\infty$ be a sequence of measures on a $\sigma$-algebra $\mathcal{A}$, and let $\{\mu_n\}$ converge to a function $\mu$. In other words, suppose that for every $A\in\mathcal{A}$, $\lim \mu_n(A) = \mu(A)$. Also, suppose that there is a finite measure $\nu$ on $\mathcal{A}$, such that for every $n\in\mathbb{N}$ and for every set $A\in\mathcal{A}$, $\mu_n(A)\le\nu(A)$. Prove that $\mu$ is a measure on $\mathcal{A}$. Does this remain true if the measure $\nu$ is not finite?
I was able to prove that $\mu$ is indeed a measure:
1.) $\mu: \mathcal{A}\longrightarrow[0,\infty]$.
Proof: Since $\mu_n$ is a measure on $\mathcal{A}$ for each $n\in\mathbb{N}$, then we must have $\mu_n:\mathcal{A}\longrightarrow[0,\infty]$ for each $n\in\mathbb{N}$. But suppose there exists some $A\in\mathcal{A}$ such that $\mu(A)\not\in[0,\infty]$, so $\mu(A)\in[-\infty,0)$. Thus there exists some $\varepsilon>0$ such that $\mu(A)+\varepsilon=0$. But since $\lim_{n\rightarrow\infty}\mu_n(A)=\mu(A)$, then there exists $N\in\mathbb{N}$ such that $|\mu(A)-\mu_N(A)|<\varepsilon$. Thus for all $n>N$ we would have $\mu_n(A)<0$, a contradiction. Thus $\mu: \mathcal{A}\longrightarrow[0,\infty]$.
2.) $\mu(\emptyset)=0$.
Proof: Since $\mu_n$ is a measure on $\mathcal{A}$ for each $n\in\mathbb{N}$, then we must have $\mu_n(\emptyset)=0$ for all $n\in\mathbb{N}$. Thus
$$
\mu(\emptyset)=\lim_{n\rightarrow\infty}\mu_n(\emptyset)=\lim_{n\rightarrow\infty}0=0.
$$
3.) $\mu\left(\bigcup_{k=1}^\infty A_k\right)=\sum_{k=1}^\infty\mu(A_k)$.
Proof: Since $\nu$ is a finite measure on $\mathcal{A}$, then for a countable collection of pairwise disjoint sets $\{A_i\}\in\mathcal{A}$ we have
$$
\nu\left(\bigcup_{k=1}^\infty A_k\right)=\sum_{i=k}^\infty\nu(A_k)<\infty.
$$
Now for each $n\in\mathbb{N}$ we have $\mu_n(A_k)<\nu(A_k)$, so the Weierstrass M-Test tells us that $\sum_{k=1}^{\infty}\mu_n(A_k)$ converges uniformly for each $n\in\mathbb{N}$.
So finally we have
$$
\mu\left(\bigcup_{k=1}^\infty A_k\right)=\lim_{n\rightarrow\infty}\mu_n\left(\bigcup_{k=1}^\infty A_k\right)\overset{(1)}{=}\lim_{n\rightarrow\infty}\sum_{k=1}^\infty\mu_n(A_i)\overset{(2)}{=}\sum_{k=1}^\infty\lim_{n\rightarrow\infty}\mu_n(A_k)=\sum_{k=1}^\infty\mu(A_k),
$$
where (1) is guaranteed since $\mu_n$ is a measure for every $n\in\mathbb{N}$ and (2) is guaranteed by the uniform convergence of $\sum_{k=1}^{\infty}\mu_n(A_k)$ for each $n\in\mathbb{N}$. Thus $\mu$ is a measure on $\mathcal{A}$.
Clearly, when $\nu$ is not a finite measure, we cannot claim uniform convergence and pass the limit inside the summation. I would like a good example that demonstrates this. Any ideas?
Best Answer
The original answer was wrong. By the Vitali-Hahn-Saks theorem, no dominating measure is needed if $\mu$ is everywhere finite.
The reason that my purported example does not work is that we do not get convergence for every measurable set. If $E$ is the set of even numbers, then $\lim_n \delta_n(E)$ does not exist.
If $\mu$ is not everywhere finite, $\mu$ need not be countably additive though. Here is a correct example: Take the underlying measurable space to be $\mathbb{N}$ endowed with the $\sigma$-algebra $2^\mathbb{N}$. Let $\mu_n(A)=\#\{m\in A\mid m>n\}$, with value $\infty$ if the set in question is infinite. If $\nu$ is counting measure, then $\mu_n(A)\leq\nu(A)$ for each $n$ and each $A\subseteq\mathbb{N}$. Then $\lim_n \mu_n(A)=0$ for $A$ finite and $\lim_n \mu_n(A)=\infty$ for $A$ infinite. The resulting set-function $\mu$ is not countably additive since $$\sum_n \mu\big(\{n\}\big)=0<\infty=\mu(\mathbb{N}).$$
Here is a simple example that shows why the assumption that all measures are dominated by some finite measure cannot simply be dropped:
Take the underlying measurable space to be $\mathbb{N}$ endowed with the $\sigma$-algebra $2^\mathbb{N}$. Let $\delta_n$ be the Dirac measure concentrated on $n$ given by $$\delta_n(A)=\begin{cases} 1 &\mbox{if } n\in A\\ 0 & \mbox{if } n\notin A. \end{cases}.$$
All the $\delta_n$ are dominated by counting measure, which is $\sigma$-finite, but they are clearly not dominated by any finite measure.
Now, $$\sum_{m\in\mathbb{N}}\lim_{n\to\infty}\delta_n\Big(\{m\}\Big)=0\neq\lim_{n\to\infty}\delta_n(\mathbb{N})=1,$$ so the limit set function is not countable additive.