# A finitely additive measure is measure if and only if continuous from below

Theorem 1.8 Let $(X,M,\mu)$ be a measure space

c.) (continuity from below) If $\{E_j\}_{1}^{\infty}\subset M$ and $E_1\subset E_2\subset\ldots$, then $\mu(\cup_{1}^{\infty}E_j) = \lim_{j\rightarrow \infty}\mu(E_j)$

A finitely additive measure $\mu$ is a measure if and only if it is continuous from below as in theorem 1.8c.

Proof: Let's assume $\mu$ is a finitely additive measure, and continuous from below. We want to show that $\mu$ is a measure. Suppose we have $\{E_j\}_{1}^{\infty}\subset M$. Set, $$F_k = E_k\setminus \Bigg[\bigcup_{1}^{k-1}E_j\Bigg] = E_k \cap \Bigg[\bigcup_{1}^{k-1}E_j\Bigg]^{c}$$ Then $F_k$'s are disjoint and belong to $M$, and $\cup_{1}^{\infty}F_k = \cup_{1}^{\infty}E_j$. By the continuity from below, $$\mu\Bigg(\bigcup_{1}^{\infty}E_j\Bigg) = \lim_{j\rightarrow\infty}\mu(E_j)$$ Since $\mu$ is finitely additive, and $E_1,\ldots,E_n$ are disjoint then $$\mu(F_k) = \mu\Bigg(\bigcup_{1}^{n}E_j\Bigg) = \sum_{1}^{n}\mu(E_j)$$ Hence $$\mu\Bigg(\bigcup_{1}^{\infty}E_j\Bigg) = \mu\Bigg(\bigcup_{1}^{\infty}F_k\Bigg) = \lim_{k\rightarrow\infty}\mu(F_k) = \lim_{n\rightarrow\infty}\sum_{1}^{n}\mu(E_j) = \sum_{1}^{\infty}\mu(E_j)$$ Therefore, $\mu$ is countably additive. If, $\mu$ is a measure, by theorem 1.8c, $\mu$ is continuous from below.

I am not sure if I am right, any suggestions would be greatly appreciated.

When you first you the continuity from below, the $E_j$ have to be an increasing sequence. You didn't assume this in the first place. After, you're writing:

$$\mu(F_k) = \mu\Bigg(\bigcup_{1}^{n}E_j\Bigg)$$. Where does this come from?

I suggest you simply define $$F_k=\bigcup_{1}^{k}E_j$$.

Clearly, $$\bigcup_{k\geq 1}F_k=\bigcup_{j\geq 1}E_j$$

Then, as $\mu$ is finitely additive, and as the $E_j$ are disjoint (you must assume this at the beginning of your proof), $$\mu\left(F_k\right)=\sum_{j=1}^k\mu(E_j)$$.

Now you can use the continuity from below, since $(F_k)$ is an increasing sequence of measurable sets:

$$\mu\left(\bigcup_{j\geq 1}E_j\right)=\mu\left(\bigcup_{k\geq 1}F_k\right)=\lim_{n\rightarrow +\infty}\mu(F_k)=\lim_{n\rightarrow +\infty}\sum_{j=1}^k\mu(E_j)=\sum_{j=1}^{+\infty}\mu(E_j)$$