Measure Theory (II): Measures

12 Sep 2018

• measure theory

Recall that we defined $\sigma$-algebras as the natural setting for defining a measure, which should capture some basic intuitive properties of the “size,” “length,” or “volume” of a set.

Formally, given a $\sigma$-algebra $\Sigma$ on a set $\Omega$ (which we think of as the collection of all measurable subsets of $\Omega$), we define a measure on $\Sigma$ to be a function $\mu$ satisfying:

• $\mu:\Sigma\to[0,\infty]$;
• $\mu(\emptyset)=0$;
• If $A_1,A_2,\ldots\in\Sigma$ are disjoint, then

$$\mu\left(\bigcup_{k=1}^\infty A_k\right)=\sum_{k=1}^\infty\mu(A_k).$$

The most important axiom here is the last one, which states the countable additivity property for measures. Just as in the definition of a $\sigma$-algebra, countable additivity implies finite additivity, by taking $A_k=\emptyset$ for all $k>N$.

Note that in order for these axioms to make sense, we need $\emptyset$ to be measurable, and countable unions of measurable sets to be measurable. This justifies two of the axioms defining a $\sigma$-algebra.

Given a set $\Omega$ and a $\sigma$-algebra $\Sigma$ on $\Omega$, we call the pair $(\Omega,\Sigma)$ a measurable space. If $\mu$ is a measure on $\Sigma$, we call the triple $(\Omega,\Sigma,\mu)$ a measure space, and any $A\in\Sigma$ is called $\Sigma$-measurable, $\mu$-measurable, or simply measurable if the measure space is clear from context.

For example, for any set $\Omega$, let $\mu:\mc P(\Omega)\to[0,\infty]$ be given by $\mu(A)=\lvert A\rvert$ if $A$ is finite, and $\mu(A)=\infty$ if $A$ is infinite. Then it is easy to check that $\mu$ is a measure on $\mc P(\Omega)$, called the counting measure on $\Omega$.

Proposition: Let $(\Omega,\Sigma,\mu)$ be a measure space, and let $A,B\in\Sigma$ with $A\subseteq B$. Then

• $\mu(A)\leq\mu(B)$;
• If $\mu(B)<\infty$, then $\mu(B\backslash A)=\mu(B)-\mu(A)$.

Proof: Both parts follow from the fact that $B$ is the disjoint union of $A$ and $B\backslash A$. $\qed$

The following result states that, when passing to the limit, the measure of nested sequences of sets behave as expected.

Proposition: Let $(\Omega,\Sigma,\mu)$ be a measure space, and let $A_1,A_2,\ldots\in\Sigma$.

1. If $A_1\subseteq A_2\subseteq\cdots$, then

   $$\mu\left(\bigcup_{k=1}^\infty A_k\right)=\lim_{k\to\infty}\mu(A_k);$$

2. If $A_1\supseteq A_2\supseteq\cdots$, and $\mu(A_{k_0})<\infty$ for some $k_0$, then

   $$\mu\left(\bigcap_{k=1}^\infty A_k\right)=\lim_{k\to\infty}\mu(A_k).$$

Proof: (1): Let $B_n=A_n\backslash A_{n-1}$. By induction, we have $A_n=\bigcup_{k=1}^nB_k$, so $B_n$ is disjoint from $B_1,\ldots,B_{n-1}$. Thus

$$\begin{aligned} \mu\left(\bigcup_{k=1}^\infty A_k\right)&=\mu\left(\bigcup_{k=1}^\infty B_k\right)\\ &=\sum_{k=1}^\infty\mu(B_k)\\ &=\lim_{n\to\infty}\sum_{k=1}^n\mu(B_k)\\ &=\lim_{n\to\infty}\mu\left(\bigcup_{k=1}^nB_k\right)\\ &=\lim_{n\to\infty}\mu(A_n). \end{aligned}$$

(2): Since $A_{k_0}\backslash A_{k_0+1}\subseteq A_{k_0}\backslash A_{k_0+2}\subseteq\cdots$, we have

$$\begin{aligned} \mu\left(\bigcap_{k=1}^\infty A_k\right)&=\mu(A_{k_0})-\mu\left(A_{k_0}\backslash\bigcap_{k=1}^\infty A_k\right)&&(\because\mu(A_{k_0})<\infty)\\ &=\mu(A_{k_0})-\mu\left(\bigcup_{k=1}^\infty A_{k_0}\backslash A_k\right)\\ &=\mu(A_{k_0})-\lim_{k\to\infty}\mu(A_{k_0}\backslash A_k)&&(\text{by }(1))\\ &=\mu(A_{k_0})-\lim_{k\to\infty}(\mu(A_{k_0})-\mu(A_k))&&(\because\mu(A_{k_0})<\infty)\\ &=\lim_{k\to\infty}\mu(A_k).\qed \end{aligned}$$

Note that the finiteness assumption in (2) is necessary: for example, if $\Omega=\bb N$, $\mu$ is the counting measure, and $A_n=\{n,n+1,\ldots\}$, then each $A_n$ has measure $\infty$, but the intersection has measure zero!

In general, we have the subadditivity property for measures:

Proposition: Let $(\Omega,\Sigma,\mu)$ be a measure space, and let $A_1,A_2,\ldots\in\Sigma$. Then

$$\mu\left(\bigcup_{k=1}^\infty A_k\right)\leq\sum_{k=1}^\infty\mu(A_k).$$

Proof: Let $B_n=A_n\backslash\bigcup_{k=1}^{n-1}A_k$. By induction, we have $\bigcup_{k=1}^nA_k=\bigcup_{k=1}^nB_k$, so $B_n$ is disjoint from $B_1,\ldots,B_{n-1}$. Thus

$$\begin{aligned} \mu\left(\bigcup_{k=1}^\infty A_k\right)&=\mu\left(\bigcup_{k=1}^\infty B_k\right)\\ &=\sum_{k=1}^\infty\mu(B_k)\\ &\leq\sum_{k=1}^\infty\mu(A_k), \end{aligned}$$

since $B_k\subseteq A_k$. $\qed$

A measure also formalises the idea of a “small,” “thin,” or “sparse” set. More precisely, given a measure space $(\Omega,\Sigma,\mu)$, a $\mu$-null set is defined as any $N\in\Sigma$ with $\mu(N)=0$. Then for any property $P$ defined on the elements of $\Omega$, we say that $P$ holds almost everywhere (or a.e.) if $P$ only fails on a $\mu$-null set.