# Finite additive measure is a measure if it is continuous from below

• September 27th 2009, 09:12 AM
Finite additive measure is a measure if it is continuous from below
Suppose that a finite additive measure $\mu : \mathbb {M} \rightarrow [0, \infty ]$ is continuous from below, I need to show that it is a measure.

Proof.

Let $\{ E_n \} ^ \infty _{n=1} \subset \mathbb {M}$, all disjoint.

Define $F_n= \bigcup ^n _{j=1}E_j$.

Note that: 1. $\bigcup _{j=1} ^ \infty F_j = \bigcup _{j=1}^ \infty E_j$ and 2. $F_j \subset F_{j+1}$

Then we have $\mu ( \bigcup _{j=1}^ \infty F_j) = \lim _{n \rightarrow \infty } \mu (F_n)$

Now, $\mu ( \bigcup _{j=1}^ \infty E_j ) = \mu ( \bigcup _{j=1}^ \infty F_j ) = \lim _{n \rightarrow \infty } \mu (F_n) = \lim _{ n \rightarrow \infty } \mu ( \bigcup _{j=1}^n E_j )$ $= \lim _{n \rightarrow \infty } \sum _{j=1}^n \mu (E_j) = \sum _{j=1}^ \infty \mu (E_j)$.

Is this correct? Thank you.
• September 27th 2009, 09:28 AM
Moo
Looks correct (Clapping)