measure

For a set X with $\sigma$ -algebra $\mathcal{S}$ on X. A measure on $(X,\mathcal{S})$ is a function $\mu :\mathcal{S}\to [0,\infty ]$ such that:

1. $\mu (\varnothing )=0$
2. $\mu \left({\bigcup }_{k=1}^{\infty }{E}_k\right)={\sum }_{k=1}^{\infty }\mu (E_k)$ for disjoint sets $E_1,E_2,...\in \mathcal{S}$

Properties

Preserves Order

For sets $A\subset B⟹\mu (A)\le \mu (B)$

Proof

Since $A\subset B$, $\exists C\subset B$ such that $A=B/C$ or $B=A\bigsqcup C⟹\mu (B)=\mu (A)+\mu (C)$ Because $\mu (C)\ge 0$ we know that $\mu (B)\ge \mu (A)$

Measure of set difference

For sets $A\subset B⟹\mu (B\backslash A)=\mu (B)-\mu (A)$

Proof

This one is pretty obvious…

Countable subadditivity

For measure space $(X,\mathcal{S},\mu )$, and $E_1,E_2,...\in \mathcal{S}$, $\mu \left({\bigcup }_{k=1}^{\infty }{E}_k\right)\le {\sum }_{k=1}^{\infty }\mu (E_k)$

Note: This is a big one. Very important theorem.

Proof

Key idea: Make the sets disjoint. Then we can use the definition of them measure. We want to construct sets $D_i$ that we can ‘subtract’ out of the other sets to make the $E_i$ sequence disjoint. Let $D_1=\varnothing$ and $D_i={\bigcup }_{k=1}^i{E}_k$ Then $E_1\cup \cdots \cup E_n=E_1\backslash D_1\cup \cdots \cup E_n\backslash D_n$ since each $D_i$ only takes out the part that has already been included in the union. Thus, $\mu \left({\bigcup }_{k=1}^{\infty }{E}_k\right)=\mu \left({\bigcup }_{k=1}^{\infty }{E}_k\backslash D_k\right)\le {\sum }_{k=1}^{\infty }\mu (E_k)$ since $\mu (D_i)\ge 0$

Continuity of measure

1. Increasing union version: For sequence of sets $E_1\subseteq E_2\subseteq \cdots \in \mathcal{S}$ (an increasing sequence of sets) $\mu \left({\bigcup }_{k=1}^{\infty }{E}_k\right)={\lim }_{k\to \infty }\mu (E_k)$

2. Decreasing intersection version: For sequence of sets $E_1\supseteq E_2\supseteq \cdots \in \mathcal{S}$ (an decreasing sequence of sets) and $\mu (E_i)<\infty$ $\mu \left({\bigcap }_{k=1}^{\infty }{E}_k\right)={\lim }_{k\to \infty }\mu (E_k)$

Proof

1. Since each set is a superset of the previous, we can ‘make’ them all disjoint. WLOG, let $E_0=\varnothing$ so the $E_1\backslash E_0=E_1$ Then ${\bigcup }_{k=1}^{\infty }{E}_k={\bigcup }_{k=1}^{\infty }{E}_k\backslash E_{k-1}$ which is a disjoint union.

Thus,

$$\begin{array}{rl}\mu \left(\bigcup _{k=1}^{\infty }{E}_k\right)& =\mu \left(\bigcup _{k=1}^{\infty }{E}_k\backslash E_{k-1}\right)\\ & =\sum _{j=1}^{\infty }\mu (E_j\backslash E_{j-1})\\ & =\underset{k\to \infty }{\lim }\sum _{j=1}^k\mu (E_j\backslash E_{j-1})\\ & =\underset{k\to \infty }{\lim }\sum _{j=1}^k\mu (E_j)-\mu (E_{j-1})\\ & =\underset{k\to \infty }{\lim }\mu (E_k)\end{array}$$

2. This one depends on De Morgan’s law and then uses 1.