Exercise 1.4.23 (Properties of countably additive measure)

In the following, we mimic the solution to the Exercise 1.2.11.

(i)

Countable subadditivity
We try to make ${\left(E\right)}_{n\in \mathbb{N}}$ disjoint to make use of the countable additivity property. For this, notice the identity $$\bigcup _{n=1}^{\infty }{E}_n=\bigcup _{n=1}^{\infty }\left[E_n∕\bigcup _{i\in \mathbb{N}∕\{n\}}{E}_i\right].$$

Using the above, we obtain

$$\mu \left(\bigcup _{n=1}^{\infty }{E}_n\right)=\mu \left(\bigcup _{n=1}^{\infty }\left[E_n∕\bigcup _{i\in \mathbb{N}∕\{n\}}{E}_i\right]\right)=\sum _{n=1}^{\infty }\mu \left(E_n∕\bigcup _{i\in \mathbb{N}∕\{n\}}{E}_i\right)\le \sum _{n=1}^{\infty }\mu \left(E_n\right).$$

(ii)

Upwards monotone convergence
Notice that we can write the union as $$\bigcup _{n=1}^{\infty }{E}_n=\bigcup _{n=1}^{\infty }{E}_n∕E_{n-1}$$

Where the right-hand side is contained in the left-hand side since for all $n\in \mathbb{N}:E_n\supseteq E_n∕E_{n-1}$, and the left-hand side is contained in the right-hand side since for any $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n$ we have $x\in E_k∕E_{k-1}$ for $k:=\min <!--\; FUNCTION\; APPLICATION\; -->\{n\in \mathbb{N}:x\in E_n\}$. Furthermore, this union is disjoint, since for any $n,m\in \mathbb{N}:n>m$ we have $(E_n∕E_{n-1})\cap (E_m∕E_{m-1})\subseteq (E_n∕E_{n-1})\cap E_m\subseteq E_m∕E_{n-1}=\varnothing$. Thus, we can invoke the countable additivity property and obtain

$$\mu \left(\bigcup _{n=1}^{\infty }{E}_n\right)=\mu \left(\bigcup _{n=1}^{\infty }{E}_n∕E_{n-1}\right)=\sum _{n=1}^{\infty }\mu (E_n)-\mu (E_{n-1})$$

By the law of telescoping series (cf. Lemma 7.2.15 from Analysis I), this series converges to $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}m(E_n)$.

(iii)

Downwards monotone convergence
We combine two facts. On one hand, since ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n\subseteq E_1$ we have $$\mu \left(E_1∕\bigcap _{n=1}^{\infty }{E}_n\right)=\mu \left(E_1\right)-\mu \left(\bigcap _{n=1}^{\infty }{E}_n\right)$$

On the other hand, note that

$$E_1∕\bigcap _{n=1}^{\infty }{E}_n=\bigcup _{n=1}^{\infty }{E}_n∕E_{n+1}$$

which implies

$$\mu \left(E_1∕\bigcap _{n=1}^{\infty }{E}_n\right)=\sum _{n=1}^{\infty }\mu \left(E_n\right)-\mu \left(E_{n+1}\right)=\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\left(\sum _{n=1}^N\mu \left(E_n\right)-\mu \left(E_{n+1}\right)\right)=\mu (E_1)-\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\mu (E_{N+1})$$

where in the last step we have used the telescoping series property again. Combining both equations we get

$$\mu \left(\bigcap _{n=1}^{\infty }{E}_n\right)=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\mu (E_n)$$