# The Unapologetic Mathematician

## Continuity of Measures

Again we start with definitions. An extended real-valued set function $\mu$ on a collection of sets $\mathcal{E}$ is “continuous from below” at a set $E\in\mathcal{E}$ if for every increasing sequence of sets $\{E_i\}\subseteq\mathcal{E}$ — that is, with each $E_i\subseteq E_{i+1}$ — for which $\lim_iE_i=E$ — remember that this limit can be construed as the infinite union of the sets in the sequence — we have $\lim_i\mu(E_i)=\mu(E)$. Similarly, $\mu$ is “continuous from above” at $E$ if for every decreasing sequence $\{E_i\}\subseteq\mathcal{E}$ for which $\lim_iE_i=E$ and which has $\lvert\mu(E_i)\rvert<\infty$ for at least one set in the sequence we have $\lim_i\mu(E_i)=\mu(E)$. Of course, as usual we say that $\mu$ is continuous from above (below) if it is continuous from above (below) at each set in its domain.

Now I say that a measure is continuous from above and below.

First, if $\{A_i\}\subseteq\mathcal{A}$ is an increasing sequence whose limit is also in $\mathcal{A}$, then $\mu(\lim_iA_i)=\lim_i\mu(A_i)$. Let’s define $A_0=\emptyset$ and calculate

\displaystyle\begin{aligned}\mu\left(\lim\limits_{i\to\infty}A_i\right)&=\mu\left(\bigcup\limits_{i=1}^\infty A_i\right)\\&=\mu\left(\biguplus\limits_{i=1}^\infty\left(A_i\setminus A_{i-1}\right)\right)\\&=\sum\limits_{i=1}^\infty\mu\left(A_i\setminus A_{i-1}\right)\\&=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\mu\left(A_i\setminus A_{i-1}\right)\\&=\lim\limits_{n\to\infty}\mu\left(\biguplus\limits_{i=1}^n\left(A_i\setminus A_{i-1}\right)\right)\\&=\lim\limits_{n\to\infty}\mu\left(A_n\right)\end{aligned}

where we’ve used countable (and finite) additivity to turn the disjoint union into a sum and back.

Next, if $\{A_i\}\subseteq\mathcal{A}$ is a decreasing sequence whose limit is also in $\mathcal{A}$, and if at least one of the $A_m$ has finite measure, then $\mu(\lim_iA_i)=\lim_i\mu(A_i)$. Indeed, if $A_m$ has finite measure then $\mu(A_n)\leq\mu(A_m)<\infty$ by monotonicity, and thus the limit must have finite measure as well. Now $\{A_m\setminus A_i\}$ is an increasing sequence, and we calculate

\displaystyle\begin{aligned}\mu(A_m)-\mu\left(\lim\limits_{i\to\infty}A_i\right)&=\mu\left(A_m\setminus\lim\limits_{i\to\infty}A_i\right)\\&=\mu\left(\lim\limits_{i\to\infty}\left(A_m\setminus A_i\right)\right)\\&=\lim\limits_{i\to\infty}\mu\left(A_m\setminus A_i\right)\\&=\lim\limits_{i\to\infty}\left(\mu(A_m)-\mu(A_i)\right)\\&=\mu(A_m)-\lim\limits_{i\to\infty}\mu(A_i)\end{aligned}

And thus a measure is continuous from above and from below.

On the other hand we have this partial converse: Let $\mu$ be a finite, non-negative, additive set function on an algebra $\mathcal{A}$. Then if $\mu$ either is continuous from below at every $A\in\mathcal{A}$ or is continuous from above at $\emptyset$, then $\mu$ is a measure. That is, either one of these continuity properties is enough to guarantee countable additivity.

Since $\mu$ is defined on an algebra, which is closed under finite unions, we can bootstrap from additivity to finite additivity. So let $\{A_i\}$ be a countably infinite sequence of pairwise disjoint sets in $\mathcal{A}$ whose (disjoint) union $A$ is also in $\mathcal{A}$, and define the two sequences in $\mathcal{A}$:

$\displaystyle B_n=\biguplus\limits_{i=1}^nE_i$
$\displaystyle C_n=A\setminus B_n$

If $\mu$ is continuous from below, $\{F_n\}$ is an increasing sequence converging to $A$. We find

\displaystyle\begin{aligned}\mu(A)&=\mu\left(\lim\limits_{n\to\infty}B_n\right)\\&=\lim\limits_{n\to\infty}\mu(B_n)\\&=\lim\limits_{n\to\infty}\mu\left(\biguplus\limits_{i=1}^nA_i\right)\\&=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\mu(A_i)\\&=\sum\limits_{i=1}^\infty\mu(A_i)\end{aligned}

On the other hand, if $\mu$ is continuous from above at $\emptyset$, then $\{C_n\}$ is a decreasing sequence converging to $\emptyset$. We find

\displaystyle\begin{aligned}\mu(A)&=\lim\limits_{n\to\infty}\mu(A)\\&=\lim\limits_{n\to\infty}\mu(B_n\uplus C_n)\\&=\lim\limits_{n\to\infty}\left(\mu(B_n)+\mu(C_n)\right)\\&=\lim\limits_{n\to\infty}\mu(B_n)+\lim\limits_{n\to\infty}\mu(C_n)\\&=\lim\limits_{n\to\infty}\mu\left(\biguplus\limits_{i=1}^nA_i\right)+\lim\limits_{n\to\infty}\mu(C_n)\\&=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\mu(A_i)+0\\&=\sum\limits_{i=1}^\infty\mu(A_i)\end{aligned}

March 23, 2010 Posted by | Analysis, Measure Theory | 18 Comments