# measure of a countable union of sets

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 $i \rightarrow \infty$?

#### Solutions Collecting From Web of "measure of a countable union of sets"

Yes.

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

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, $$\mu(\bigcup_{j=1}^{\infty}E_j)\leq \lim_{j\to\infty}\mu(E_j).$$

If $\mu(E_k)=\infty$ for some $k$ then so it is for all $n>k$, then by monotonicity $$\lim\limits_{n\to\infty} \mu(E_n) = \infty = \mu\left(\bigcup\limits_{n=1}^\infty E_n\right).$$
If $\mu(E_k) < \infty$ for all $k$, let $\{G_k\}_{k=1}^\infty$ the family of disjoints sets of $\beta$ defined by $E_0=\emptyset$ and $G_k=E_k-E_{k-1}$ for all $k\geqslant 2$ and note that $$\bigcup\limits_{k=1}^\infty G_k = \bigcup\limits_{k=1}^\infty E_k.$$ Then by $\sigma-$additivity of $\mu$ we have \begin{align*}
\mu\left(\bigcup\limits_{n=1}^\infty E_n\right) &= \mu\left(\bigcup\limits_{n=1}^\infty G_n\right) = \sum\limits_{n=1}^\infty \mu(G_n) \\ &= \sum\limits_{n=1}^\infty [\mu(E_n)-\mu(E_{n-1})] \\ &= \lim\limits_{n \to \infty} \sum\limits_{k=1}^n [\mu(E_k)-\mu(E_{k-1})] = \lim\limits_{n \to \infty} \mu(E_n).
\end{align*}