$\sigma$-additivity of the limit Let $(A,\mathcal{F})$ be a measurable space. On this space, define $\{\mu_n\}_{n=1}^\infty$ bounded sequence so that $\lim \mu_n=\mu$ (i.e. $\mu(E)=\lim \mu_n(E),\forall E\in \mathcal{F}$). Show that $\mu$ is a measure.
I could handle to show that $\mu(\emptyset)=0$ and $\mu\geq 0$.
I am struggling to show the $\sigma$-additivity.
Let $A_1,A_2,\dotsc \in \mathcal{F}$ be pairwise disjoint sets. Then
$$\mu(\cup_{k=1}^\infty A_k)=\lim_{n\rightarrow \infty}\mu_n (\cup_{k=1}^\infty A_k )=\lim_{n\rightarrow \infty}\sum_{k=1}^\infty \mu_n(A_k ).$$
I have to show that $\lim_{n\rightarrow \infty}\sum_{k=1}^\infty \mu_n(A_k )=\sum_{k=1}^\infty \lim_{n\rightarrow \infty}\mu_n(A_k )$. 
I think it has something to do with convergence of series. How to proceed?

Half of the answers are saying that the statement is true and the other half that it is false. I'm a bit confused now.
Thank you! :)
 A: You cannot prove this easily from definitions. It is an easy consequence if Vitali -Hahn -Saks Theorem. [ https://en.wikipedia.org/wiki/Vitali%E2%80%93Hahn%E2%80%93Saks_theorem ]
This theorem assumes that the measures $\mu_n$ are absolutely continuous w.r.t. some finite measure $m$ but this is always true: we can take $m (E)= \sum\limits_{k=1}^{\infty}
 \frac {\mu_n(E)} {2^{n}\mu_n(\Omega)}$. 
A: Some properties are immediately established:


*

*$\mu(\varnothing) = \lim_{n\to\infty} \mu_n(\varnothing) = 0$.

*$\mu$ is positive, i.e., for each $E \in \mathcal{F}$, we have $\mu(E) = \lim_{n\to\infty} \mu_n(E) \geq 0$.

*$\mu$ is additive, i.e., for each disjoint $E_1, \cdots, E_k \in \mathcal{F}$, we have
$$ \mu(\cup_{i=1}^{k} E_i)
= \lim_{n\to\infty} \mu_n(\cup_{i=1}^{k} E_i)
= \lim_{n\to\infty} \sum_{i=1}^{k} \mu_n(E_i)
= \sum_{i=1}^{k} \lim_{n\to\infty} \mu_n(E_i)
= \sum_{i=1}^{k} \mu(E_i). $$

*$\mu$ is super-additive, i.e., for each disjoint $E_1, E_2, \cdots \in \mathcal{F}$,
$$ \mu(\cup_{i=1}^{\infty} E_i) \geq \lim_{k\to\infty} \mu(\cup_{i=1}^{k} E_i) = \sum_{i=1}^{\infty} \mu(E_i). $$
In order to show that $\mu$ is a measure, it suffices to show $\sigma$-additivity of $\mu$. Before delving into the proof, we examine a simple non-example to find out what can go wrong.

Non-Example. Let $(A, \mathcal{F}) = (\mathbb{N}, 2^{\mathbb{N}})$ and $\mu_n = \delta_n$. Then with $E_i = \{i\}$, we know that
$$ \lim_{n\to\infty} \mu_n(\cup_{i=1}^{\infty} E_i) = 1 > 0 = \sum_{i=1}^{\infty} \lim_{n\to\infty} \mu_n(E_i). $$
In this case, set $E = E_1 \cup E_3 \cup E_5 \cup \cdots = \{1, 3, 5, \cdots\}$. Then
$$ \mu_n(E) = \begin{cases} 1, & \text{$n$ is odd} \\ 0, & \text{$n$ is even} \end{cases} $$
and so, $\{\mu_n(E)\}_{n\geq 1}$ does not converge.

What we can learn from this is as follows: The fact that $\mu$ can be strictly super-additive implies that some mass is escaping towards the infinity. By carefully tracking that mass escape, we may tinker a set $E$ so that $\{\mu_n(E)\}_{n\geq 1}$ does not converge. Let us implement this idea to the general setting.

Claim. $\mu$ is $\sigma$-additive, and hence, a measure.

Assume otherwise that $\mu$ is not $\sigma$-additive. In view of the super-additivity of $\mu$, this implies that there exists a sequence of mutually disjoint sets $\{E_i\}_{i\geq 1}$ in $\mathcal{F}$ such that
$$\mu(\cup_{i=1}^{\infty} E_i) > \sum_{i=1}^{\infty} \mu(E_i). \tag{*}$$
Our aim is to derive a contradiction out of this. Write $\delta = \mu(\cup_{i=1}^{\infty} E_i) - \sum_{i=1}^{\infty} \mu(E_i)$. (Notice that the boundedness of $\{\mu_n\}_{n\geq1}$, in whatever sense, implies that $\mu$ is also bounded, hence $\delta$ is finite.) By the additivity of $\mu$, we have
$$ \delta = \mu(\cup_{i \geq k} E_i) - \sum_{i\geq k} \mu(E_i) \quad \text{for any} \quad k \geq 1. $$
By the assumption $\text{(*)}$, we know that $\delta > 0$. Using this, we construct a set $E$ using the following algorithm:


*

*Write $T_k = \cup_{i \geq k} E_i$ for simplicity. (The letter $T$ stands for 'tail'.)

*Set $N_0 = 1$ and $K_0 = 1$ for simplicity.

*Let $j \geq 1$ and assume that $N_{j-1}$ and $K_{j-1}$ are defined.
First, from the condition $\lim_{k\to\infty} \mu_{N_{j-1}}(T_k) = 0$, we know that there exists $K_j > K_{j-1}$ such that
$$\mu_{N_{j-1}}(T_{K_j}) < \tfrac{1}{10}\delta.$$
Then, since $\lim_{n\to\infty} \mu_n(E_i) = \mu(E_i)$ for each $i$ and $\lim_{n\to\infty} \mu_n(T_{K_j}) = \mu(T_{K_j}) \geq \delta$, there exists $N_j > N_{j-1}$ such that
$$
|\mu_n(E_i) - \mu(E_i)| \leq \frac{\delta}{10K_j} \quad \text{for all} \quad i \leq K_j \quad \text{and} \quad
\mu_n(T_{K_j}) > \tfrac{9}{10}\delta
$$
holds for any $n \geq N_j$.

*Now that $\{N_j\}_{j\geq 1}$ and $\{K_j\}_{j \geq 1}$ are defined, define $E$ by
$$ E = \cup_{j\geq 1} (T_{K_{2j}}\setminus T_{K_{2j+1}}). $$
We claim that $\mu_n(E)$ does not converge, thus yielding the desired contradiction. Indeed, for each $j \geq 2$,
\begin{align*}
\mu_{N_{2j}}(E)
&\geq \mu_{N_{2j}}(T_{K_{2j}}\setminus T_{K_{2j+1}}) + \sum_{k=1}^{j-1} \mu_{N_{2j}}(T_{K_{2k}}\setminus T_{K_{2k+1}}) \\
&\geq \tfrac{9}{10}\delta - \tfrac{1}{10}\delta + \sum_{k=1}^{j-1}  \left(\mu(T_{K_{2k}}\setminus T_{K_{2k+1}})- \frac{\delta}{10K_j}(K_{2k+1}-K_{2k}) \right) \\
&\geq \tfrac{8}{10}\delta + \left(\sum_{k=1}^{j-1} \mu(T_{K_{2k}}\setminus T_{K_{2k+1}}) \right) - \tfrac{1}{10}\delta \\
&= \tfrac{7}{10}\delta + \sum_{k=1}^{j-1} \mu(T_{K_{2k}}\setminus T_{K_{2k+1}}),
\end{align*}
while
\begin{align*}
\mu_{N_{2j-1}}(E)
&\leq \mu_{N_{2j-1}}(T_{K_{2j}}) + \sum_{k=1}^{j-1} \mu_{N_{2j-1}}(T_{K_{2k}}\setminus T_{K_{2k+1}}) \\
&\leq \tfrac{1}{10}\delta + \sum_{k=1}^{j-1}  \left(\mu(T_{K_{2k}}\setminus T_{K_{2k+1}}) + \frac{\delta}{10K_j}(K_{2k+1}-K_{2k}) \right) \\
&\leq \tfrac{1}{10}\delta + \left( \sum_{k=1}^{j-1} \mu(T_{K_{2k}}\setminus T_{K_{2k+1}}) \right) + \tfrac{1}{10}\delta \\
&\leq \tfrac{2}{10}\delta + \sum_{k=1}^{j-1} \mu(T_{K_{2k}}\setminus T_{K_{2k+1}}).
\end{align*}
This tells that $|\mu_{N_{2j}}(E) - \mu_{N_{2j-1}}(E)| \geq \tfrac{1}{2}\delta$ for all $j \geq 2$, and so, $\{\mu_n(E)\}_{n\geq 1}$ does not converge as desired. ////
A: No, uniform convergence for sequences does not imply that the sum of the limit is the limit of the sum. That's true for integrals on finite measure spaces, but the sum of a sequence is the integral wrt counting measure on $\Bbb N$, an infinite measure. To be specific, let $$f_n(j)=\begin{cases}\frac1n,&(1\le j\le n),\\0,&(j>n).\end{cases}$$Then $f_n\to0$ uniformly but $\sum_jf_n(j)=1$.
