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Let $(X,\beta,\mu)$ be a measure space. Given $E_1 \subset E_2 \subset ...$, a sequence of sets in $\beta$, obtain $\mu \left( \bigcup _{i=1}^{\infty} E_i \right)$ rigorously.

Isn't this equal to the limit of $\mu (E_i)$ as $i \rightarrow \infty$?

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First, is this really an arbitrary union, or just a countable one? Second: sure, but it looks like the crux of this one is in the "rigorously," not in making the right guess. At the very least we'd better say something about the existence of the limit. –  Kevin Carlson Aug 5 '12 at 0:58
    
It is a countable union. I had another thing on my mind when I typed `arbitrary'. –  lucas Aug 5 '12 at 1:21

2 Answers 2

up vote 2 down vote accepted

Yes.

The hint is to use countable additivity and the disjoint collection of sets $(E_{i + 1} - E_i)_{i = 1}^\infty$.

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For any given positive integer $N$, $E_N\subseteq \bigcup_{j=1}^{\infty}E_i$. Monotonicity implies $\mu(E_N)\leq \mu(\bigcup_{j=1}^{\infty}E_i)$. Taking the limit, we will have $\lim_{j\to\infty}\mu(E_j)\leq \mu(\bigcup_{j=1}^{\infty}E_j)$. Thus, $$\lim_{j\to\infty}\mu(E_j)\leq \mu(\bigcup_{j=1}^{\infty}E_j).$$

For any given positive integer $N$,$\bigcup_{j=1}^{N}E_j= E_N$. We have $\mu(\bigcup_{j=1}^{N}E_j)=\mu(E_N)$ due to monotonicity. Take the limit, we get $\mu(\bigcup_{j=1}^{\infty}E_j)\leq \lim_{j\to\infty}\mu(E_j)$. Thus, $$\mu(\bigcup_{j=1}^{\infty}E_j)\leq \lim_{j\to\infty}\mu(E_j).$$

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