Let $\mathcal{M}$ be a $\sigma$-algebra of subsets of a set $X$ and let $\mu:M\to[0,\infty)$ be finitely additive, with $\mu(X)<\infty$. Prove that $\mu$ is a measure if and only if every ascending sequence of sets $\{A_k\}_{k=1}^\infty$ in $M$ satisfies continuity of measure $\mu\left(\bigcup_{k=1}^{\infty}A_k\right)=\lim_{k\to\infty}\mu(A_k)$.

I have already proved one direction of this proposition:

Let $\mathcal{M}$ be a $\sigma$-algebra of subsets of a set $X$ and let $\mu:M\to[0,\infty)$ be finitely additive with $\mu(X)<\infty$. First, let $\mu$ be a measure, and we want to see that continuity of measure of an ascending sequence of measurable sets is satisfied. Define $A_0=\emptyset$ and $C_k=A_k\setminus A_{k-1}$ for all $k\geq 1$. Because the sequence $\{A_k\}_{k=1}^\infty$ is ascending, $\{C_k\}_{k=1}^\infty$ is disjoint and $\bigcup_{k=1}^\infty A_k=\bigcup_{k=1}^{\infty}C_k$. By the countable additivity of $\mu$,

$$\mu\left(\bigcup_{k=1}^{\infty}A_k\right)=\mu\left(\bigcup_{k=1}^{\infty}C_k\right)=\sum_{k=1}^{\infty}\mu(A_k\setminus A_{k-1}.)$$

Because $\{A_k\}_{k=1}^\infty$ is ascending, excision tells us that

$$\sum_{k=1}^\infty \mu(A_k\setminus A_{k-1})=\sum_{k=1}^\infty (\mu(A_k)-\mu(A_{k-1}))$$



But I am confused about how to use the continuity of measure to show that $\mu$ is a measure. I think I must show that $\mu(\emptyset)=0$ and countable additivity (if $\{A_k\}_{k=1}^\infty$ is a measurable and disjoint collection of sets, then $\mu\left(\bigcup_{k=1}^\infty A_k\right)=\sum_{k=1}^\infty\mu(A_k)$)? But I do not see how these follow from the given statement.


Since $\mu$ is additive and $\mu(X) < \infty$, you have that $\mu(X) = \mu(X\cup\emptyset) = \mu(X) + \mu(\emptyset)$, hence $\mu(\emptyset) = 0$.

Let us prove that $\mu$ is countably additive. Let $(A_k)\subset \mathcal{M}$ be a sequence of pairwise disjoint measurable sets, and define $$ B_n := \bigcup_{k=1}^n A_k, \qquad A:= \bigcup_{k=1}^\infty A_k = \bigcup_{k=1}^\infty B_k. $$ Then, by assumption and finite additivity, $$ \mu(A) = \lim_n \mu(B_n) = \lim_n \sum_{j=1}^n \mu(A_j) = \sum_{j=1}^\infty \mu(A_j). $$

  • $\begingroup$ In the very last step, do you mean to say $\sum_{j=1}^\infty\mu(A_j)$? $\endgroup$ – MathStudent1324 Nov 14 '17 at 18:51
  • $\begingroup$ Yes, sure. Now I'll edit the answer. $\endgroup$ – Rigel Nov 14 '17 at 18:59

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