The Unapologetic Mathematician

Mathematics for the interested outsider

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

   

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