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

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

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

is a measure if and only if

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

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 $\iff$ $\mu$ is a measure.

Proof

From page 24 - although not hard to do without.

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TODO: Clean up and prove

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