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Let $(X,\Sigma,\mu)$ be a measurable space and let $(\mu_n)_{n\geq 1}$ be a sequence of measure in this space. Suppose that this is a monotone increasing sequence, in the sense that $\mu_n(E)\leq \mu_{n+1}(E)$ for all $E \in \Sigma$, and for all $n$. If we define, for every $E \in \Sigma$

$\space$ $\mu(E)=\lim_{n \to \infty} \mu_n(E)$,

prove that $\mu$ is a measure.

The attempt at a solution

I am having problems trying to prove the countable additivity property, which is, if $\{E_i\}_{i \in \mathbb N}$ is a countable collection of pairwise disjoint sets in $\Sigma$, then $\mu(\bigcup_{i \in \mathbb N} E_i)=\sum_{i \in \mathbb N} \mu(E_i)$

In this particular case, the property is satisfied if and only if

$\lim_{n \to \infty} \mu_n(\bigcup_{i \in \mathbb N} E_i)=\sum_{i \in \mathbb N}\lim_{n \to \infty}\mu_n(E_i)$

Each $\mu_n$ is a measure, so

$\lim_{n \to \infty} \mu_n(\bigcup_{i \in \mathbb N} E_i)=\lim_{n \to \infty} \sum_{i \in \mathbb N} \mu_n(E_i)$.

How could I go on from here? I know I must use the fact that $(\mu_n)_{n \in \mathbb N}$ is a monotone increasing sequence, I would appreciate any help.

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up vote 1 down vote accepted

You can use that a set function $\mu$ defined on a measurable space such that $\mu(\emptyset)=0$, $\mu$ is finitely additive and is continuous from below is a measure on that space. Also, you can define a double sequence $(s_{n,m})_{n,m\geq 1}$ given by $s_{n,m}=\sum_{k=1}^m \mu_n(E_k)=\mu_n(\bigcup_{k=1}^m E_k)$ and then show that $$\lim\limits_{n\to\infty}\lim\limits_{m\to\infty} s_{n,m}=\sup\limits_n\sup\limits_m(s_{n,m})=\sup\limits_m\sup\limits_n(s_{n,m})=\lim\limits_{m\to\infty}\lim\limits_{n\to\infty} s_{n,k}$$ Note that the equalities between the supremum and the limit is because in each index, the sequence $(s_{n,k})$ is monotonically increasing.

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