# A function is a measure iff it measures the empty set as 0, disjoint sets add, and it is continuous from below (with equiv. conditions)

## Statement

$\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }$$\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}$$\require{AMScd}\newcommand{\d}[]{\mathrm{d}^{#1} }$Let [ilmath](X,\mathcal{A})[/ilmath] be a measurable space. A map:

• [ilmath]\mu:\mathcal{A}\rightarrow[0,\infty][/ilmath]

is a measure if and only if

1. [ilmath]\mu(\emptyset)=0[/ilmath]
2. [ilmath]\mu(A\udot B)=\mu(A)+\mu(B)[/ilmath]
3. Either:
1. For any increasing sequence of sets[Note 1] [ilmath](A_n)_{n=1}^\infty\subseteq\mathcal{A}[/ilmath] with [ilmath]\lim_{n\rightarrow\infty}(A_n)=A\in\mathcal{A}[/ilmath] we have
• $\mu(A)=\lim_{n\rightarrow\infty}(\mu(A_n))=\inf_{n\in\mathbb{N} }(\mu(A_n))$
• This is called Continuity of measures from below
2. Or [ilmath]\forall A\in\mathcal{A} [/ilmath] we have [ilmath]\mu(A)<\infty[/ilmath] AND:
1. Either (these are equivalent)[Note 2]
1. For any decreasing sequence of sets[Note 3] [ilmath](A_n)_{n=1}^\infty\subseteq\mathcal{A}[/ilmath] with [ilmath]\lim_{n\rightarrow\infty}(A_n)=A\in\mathcal{A}[/ilmath] we have
• $\mu(A)=\lim_{n\rightarrow\infty}(\mu(A_n))=\inf_{n\in\mathbb{N} }(\mu(A_n))$
• This is called Continuity of measures from above
2. For any decreasing sequence of sets [ilmath](A_n)_{n=1}^\infty[/ilmath] with [ilmath]\lim_{n\rightarrow\infty}(A_n)=\emptyset[/ilmath] we have:
• $\lim_{n\rightarrow\infty}(\mu(A_n))=0$
• This is called continuity of measures at [ilmath]\emptyset[/ilmath]

## Page notes

This is actually several theorems rolled into one. Halmos has some good terminology and splits these theorems up. I will come back to this when I've done that.

As it stands now this is a good theorem with some extra facts bolted on. I like conditions 1 2 and 3.1 [ilmath]\iff[/ilmath] [ilmath]\mu[/ilmath] is a measure.

## Proof

From page 24 - although not hard to do without.

TODO: Clean up and prove