Convergence of measure sequences bounded by a finite measure

Solution 1:

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.