# 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 »

1. […] sets on which and agree, then the limit of this sequence is again in . Indeed, since measures are continuous, we must […]

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2. […] since is continuous, we see […]

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3. […] in Mean First off we want to introduce another notion of continuity for set functions. We recall that a set function on a class is continuous from above at if for every decreasing sequence of […]

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4. […] so for all . A.e. convergence tells us that the measure of the intersection of all the is . By continuity, we conclude […]

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5. […] we turn to some continuity properties. If is a monotone sequence — if it’s decreasing we also ask that at least […]

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6. […] a monotone sequence of simple so that or . Then the limit will commute with (since measures are continuous), and it will commute with the integral as […]

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7. […] what we know about continuity, we just have to show that is continuous from above at to show that it’s a measure. That […]

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8. […] we need to check is continuity. We know that it suffices to show that is continuous from above at . So, let be a decreasing sequence of measurable sets […]

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9. Very informative. Thanks!

Comment by Tim Fortune | February 8, 2013 | Reply

10. Is there a way to find a counterexample for the monotonically decreasing sequence result, for \mu(A_1) = \infty?

Comment by Alice | April 6, 2013 | Reply

11. I think you’re slightly confused, Alice, which is fine; the condition is confusing. It states that, given a set $E\in\mathcal{E}$, a certain property must hold for all decreasing sequences starting from $E_1\in\mathcal{E}$ with finite measure that converge to $E$. It doesn’t say anything about what must happen when a sequence that decreases to $E$ starts with a set of infinite measure; the property may or may not hold for such sequences, but that has no bearing on the continuity of the measure at $E$.

Comment by John Armstrong | April 6, 2013 | Reply

12. Oh right, I think I mis-read this. I think I’m speaking of the property of the monotonicity of measure whereby the measure of the infinite intersection taken over a decreasing sequence is the limit of the measure of the set $A_n$,

\mu ( \cap^{\infty} A_n)= \lim \mu (A_n), where A_{n+1} \subset A_n.

So here, I’m asking for an example where the sequence starts with infinite measure, and this condition fails to hold. So sorry for this confusion!

Comment by Alice | April 6, 2013 | Reply

13. Well, I could imagine cooking up a space with a countably infinite number of points assigned an infinite measure each; the sequence is $E$ along with all of them, and then removing one point at a time. The limit of the sequence is $E$ since each point will eventually be removed, but the measure of each set is infinite.

I’m not really a measure theorist so I’m sort of waving my hands here at the idea that such a pathological space would be valid, but I think something along those lines should work.

Comment by John Armstrong | April 6, 2013 | Reply

14. Would the Lebesgue measure on [n, \infty ) work?

Comment by Alice | April 7, 2013 | Reply

15. Well, the trick is to get it so that the limit of the sequence has a finite measure. Otherwise the limit of the measures is the measure of the limit. I suppose in this case the limit is empty, so that might qualify.

Comment by John Armstrong | April 7, 2013 | Reply

16. well the limit on the measure would be 0 which is definitely finite, I’m slightly dubious that the measure of the infinite intersection may not be different though….?

Comment by Alice | April 7, 2013 | Reply

17. No, each of those sets has infinite measure, so the limit of the measures is infinite. But the intersection is empty (no real number can be in it), so the measure of the limit is zero.

Comment by John Armstrong | April 8, 2013 | Reply

18. I see, thank you so much for clearing this up!

Comment by Alice | April 8, 2013 | Reply