### A Mild Glimpse Into Measure Theory (2)

3. Measures and Outer measures

Let ${(E,\mathscr{M})}$ be a measurable space.

Definition: A measure on ${(E,\mathscr{M})}$ is a function ${\mu:\mathscr{M}\rightarrow [0,\infty]}$ such that:
(1) ${\mu(\emptyset)=0}$
(2) For all families ${\left\{A_{n}\right\}_{n\in\mathbb{N}}}$ of disjoint measurable sets,

$\displaystyle \mu(\bigcup_{n\in\mathbb{N}} A_{n})=\sum_{n\in\mathbb{N}}\mu(A_{n})$

Remarks:

• ${\mu(A)=+\infty}$ is allowed
• Property (2) is called ${\sigma}$-additivity
• We can also derive ${\mu(\bigcup_{n=1}^{k} A_{n})=\sum_{n=1}^{k}\mu(A_{n})}$

Proposition 6 : Let ${A,B\in\mathscr{M}}$, we have the following:
(1) If ${A\subseteq B}$, then ${\mu(A)\leq\mu(B)}$.
(2) If ${\mu(A)<+\infty}$ and ${A\subseteq B}$, then ${\mu(A\setminus B)=\mu(A)-\mu(B)}$
(3) ${\mu(A\cup B)=\mu(A)+\mu(B)-\mu(A\cap B)}$
(4) If ${A_{n}\in\mathscr{M}}$ for all ${n\in\mathbb{N}}$ and ${A_{n}\subseteq A_{n+1}}$ for all ${n\in\mathbb{N}}$, then ${\mu(\bigcup_{n\in\mathbb{N}}A_{n})=\lim_{n\rightarrow\infty}\mu(A_{n})}$.
(5) If ${B_{n}\in\mathscr{M}}$ for all ${n\in\mathbb{N}}$ and ${B_{n}\supseteq B_{n+1}}$ for all ${n\in\mathbb{N}}$, then ${\mu(\bigcap_{n\in\mathbb{N}}B_{n})=\lim_{n\rightarrow\infty}\mu(B_{n})}$.
(6) If ${A_{n}\in\mathscr{M}}$, then ${\mu(\bigcup_{n\in\mathbb{N}}A_{n})\leq\sum_{n\in\mathbb{N}}A_{n}}$.

To be updated.