# Prove that $\mu$ is a measure under several conditions.

Suppose that $${\cal F}$$ is a $$\sigma$$-algebra on a set $$X$$ and $$\mu\mathop:{\cal F} \to [0,\infty]$$ satisfies the conditions:

1. $$\mu(\emptyset) = 0$$.
2. For every pair $$A$$ and $$B$$ of disjoint sets in $${\cal F}$$, $$\mu(A \cup B) = \mu(A) + \mu(B)$$.
3. For every decreasing sequence $$\{E_n\}$$ in $${\cal F}$$ (that is $$E_{n+1} \subseteq E_n$$ for all $$n$$) such that $${\bigcap_{n =1}^{\infty} E_n = \emptyset}$$, we have $$\lim_{n \to \infty} \mu(E_n) = 0$$.

Prove that $$\mu$$ is a measure on $${\cal F}$$.

Here's my attempt:

Proof.

Let $$\{E_n\}$$ be a countably infinite collection of sets such that $$E_i \cap E_j =\emptyset$$ for all $$i,j$$. Write $$E = \bigcup_{n=1}^\infty{E_n}$$ and let $$F_n = E \setminus\bigcup_{k=1}^n{E_k}.$$ for $$n\geq 1.$$ Then we have $$F_{n+1}= E \setminus\bigcup_{k=1}^{n+1}{E_k} \subseteq E \setminus\bigcup_{k=1}^n{E_k}=F_n$$ and $$\bigcap_{n=1}^\infty{F_n} = \emptyset.$$ Hence, by applying condition (2), we have \begin{align*}\mu\left(F_n\right) & =\mu\left(E \setminus\bigcup_{k=1}^n{E_k}\right)\\ & = \mu(E) - \mu\left(\bigcup_{k=1}^n{E_k}\right)\\ & = \mu(E) - \sum_{k=1}^n{E_k} \end{align*} and the above holds for all $$n\in \mathbb{N}$$. Thus, applying condition (3), we have \begin{align*}\mu(E) & = \lim_{n\to \infty}\mu(F_n) + \lim_{n\to \infty}\sum_{k=1}^n{E_k}\\ & = \sum_{k=1}^\infty{E_k}. \end{align*} This shows that $$\mu$$ is a measure.

• I haven't made any progress. I know that it is sufficient to show that for any countably infinite collection $\{E_n\}$ of pairwise disjoint sets in $\cal{F}$, we have $\mu(\bigcup_{n=1}^\infty{E_n}) = \sum_{n=1}^\infty{\mu(E_n})$. I am not sure how the other hypotheses fit into this. I think I need to write $\bigcup_{n=1}^\infty{E_n}$ as a sequence satisfying property (3). – johnny133253 Feb 6 '19 at 6:53
• In order to apply the conditions, you need to build a decreasing sequence of sets. How can you build a decreasing sequence of sets based on the sets $E_n, \bigcup_{n} E_n$? – supinf Feb 6 '19 at 6:57

There are some minor typos in some places, where you wrote $$E_k$$ instead of $$\mu(E_k)$$, but I am sure you meant the right thing.