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.

Answer

Your proof is correct.

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.