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Let $(X,A)$ be a measurable space. Show that if $(\mu_n)$ is an increasing sequence of measures, then $\mu(A) = \lim_{n\rightarrow \infty} \mu_n(A)$ defines a measure on $(X,A)$.

i) $\mu(\emptyset) = 0$ since its true for all $\mu_n$.

ii) $\mu (\cup u_i) = \lim_{n\rightarrow \infty} \sum_i \mu_n(u_i)$ by definition. But how can I move the limes inside?

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2 Answers 2

up vote 4 down vote accepted

Hint

$$\lim_{n \to \infty} \sum_{i=1}^{\infty} \mu_n(u_i) = \sup_{n \in \mathbb{N}} \sup_{k \in \mathbb{N}} \sum_{i=1}^k \mu_n(u_i) = \sup_{k \in \mathbb{N}} \sup_{n \in \mathbb{N}} \sum_{i=1}^k \mu_n(u_i) = \ldots$$

The first equality follows from the fact that the sequence of measures is increasing (i.e. $\lim = \sup$).

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U need to use the following claim, which is in essence the same as answer by saz.

If $a_{mn}$ in monotone increasing in both m and n, then $$\lim_{m\to\infty}\lim_{n\to\infty} a_{mn}=\lim_{n\to\infty}\lim_{m\to\infty}a_{mn}$$

Using this you will get $$\lim_{n\to \infty}\lim_{m\to \infty} \sum_{i=1}^m \mu_n(u_i) = \lim_{m\to \infty}\lim_{n\to \infty} \sum_{i=1}^m \mu_n(u_i)=\lim_{m\to \infty} \sum_{i=1}^{m} \mu(u_i)$$ $$=\sum_{i=1}^{\infty} \mu(u_i)$$

Proof of claim: Let $M=\sup_{m\geq 1}\sup_{n\geq 1} a_{mn}$. Then $$a_{mn} \leq M \Rightarrow \sup_{n\geq 1}\sup_{m\geq 1} a_{mn} \leq M = \sup_{m\geq 1}\sup_{n\geq 1} a_{mn}$$ Now let $N=\sup_{n\geq 1}\sup_{m\geq 1} a_{mn}$. Similarly $$a_{mn} \leq N \Rightarrow \sup_{m\geq 1}\sup_{n\geq 1} a_{mn} \leq N = \sup_{n\geq 1}\sup_{m\geq 1} a_{mn}$$ Thus M=N and we are done.

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