# The Unapologetic Mathematician

## Measures and Induced Measures

If we start with a measure $\mu$ on a ring $\mathcal{R}$, we can extend it to an outer measure $\mu^*$ on the hereditary $\sigma$-ring $\mathcal{H}(\mathcal{R})$. And then we can restrict this outer measure to get an actual measure $\bar{\mu}$ on the $\sigma$-ring $\overline{\mathcal{S}}$ of $\mu^*$-measurable sets. And so we ask: how does the measure $\mu$ relate to the measure $\bar{\mu}$.

First, we will show that any set $E\in\mathcal{R}$ is $\mu^*$-measurable. Then, since $\overline{\mathcal{S}}$ is a $\sigma$-ring containing $\mathcal{R}$ it will also contain the smallest such $\sigma$-ring $\mathcal{S}(\mathcal{R})$.

So, if $E\in\mathcal{R}$ and $A\in\mathcal{H}(\mathcal{R})$, and if $\epsilon>0$, then the definition of the induced outer measure $\mu^*$ tells us that there exists a sequence $\{E_i\}_{i=1}^\infty\subseteq\mathcal{R}$ so that $A$ is contained in their union and

\displaystyle\begin{aligned}\mu^*(A)+\epsilon&\geq\sum\limits_{i=1}^\infty\mu(E_i)\\&=\sum\limits_{i=1}^\infty\left(\mu(E_i\cap E)+\mu(E_i\cap E^c)\right)\\&\geq\mu^*(A\cap E)+\mu^*(A\cap E^c)\end{aligned}

Since this holds for all $\epsilon$, we conclude that $\mu^*(A)\geq\mu^*(A\cap E)+\mu^*(A\cap E^c)$, and thus that $E$ is $\mu^*$-measurable.

Now $\bar{\mu}$ is defined for every set in $\mathcal{S}(\mathcal{R})$. And since it’s a measure on each of the rings $\mathcal{S}(\mathcal{R})$ and $\overline{\mathcal{S}}$, we can induce outer measures from each! But since these are $\sigma$-rings, life is a little easier. For one thing, the hereditary $\sigma$-ring each one generates is just $\mathcal{H}(\mathcal{R})$ again; for another, the induced outer measure of a set $E$ is just the infimum of the measure of any set containing $E$. And, as it turns out, both of the induced outer measures are exactly $\mu^*$! That is, for any $E\in\mathcal{H}(\mathcal{R})$ we find:

\displaystyle\begin{aligned}\mu^*(E)&=\inf\left\{\bar{\mu}(F)\vert E\subseteq F\in\mathcal{S}(\mathcal{R})\right\}\\&=\inf\left\{\bar{\mu}(F)\vert E\subseteq F\in\overline{\mathcal{S}}\right\}\end{aligned}

Indeed, by definition we have

\displaystyle\begin{aligned}\mu^*(E)&=\inf\left\{\sum\limits_{i=1}^\infty\mu(E_i)\bigg\vert E\subseteq\bigcup\limits_{i=1}^\infty E_i,\{E_i\}\subseteq\mathcal{R}\right\}\\&\geq\inf\left\{\sum\limits_{i=1}^\infty\bar{\mu}(E_i)\bigg\vert E\subseteq\bigcup\limits_{i=1}^\infty E_i,\{E_i\}\subseteq\mathcal{S}(\mathcal{R})\right\}\\&=\inf\left\{\bar{\mu}(F)\vert E\subseteq F\in\mathcal{S}(\mathcal{R})\right\}\\&\geq\inf\left\{\bar{\mu}(F)\vert E\subseteq F\in\overline{\mathcal{S}}\right\}\end{aligned}

since everything in the first set is also in the second, and everything in the third set is also in the fourth. The equality in the middle holds because every sequence in $\mathcal{S}(\mathcal{R})$ can be replaced by a disjoint sequence, and the sum of the measures can then be replaced by the measure of the disjoint union, and so we only ever need to use one set in $\mathcal{S}(\mathcal{R})$.

But since $\bar{\mu}(F)=\mu^*(F)$ for every $F\in\overline{\mathcal{S}}$, we must have $\inf\left\{\bar{\mu}(F)\vert E\subseteq F\in\overline{\mathcal{S}}\right\}\geq\mu^*(E)$. Thus all the inequalities above are equalities, as we claimed.

March 31, 2010 Posted by | Analysis, Measure Theory | 2 Comments

## Sets Measurable by an Outer Measure II

Yesterday, we showed that — given an outer measure $\mu^*$ on a hereditary $\sigma$-ring $\mathcal{H}$ — the collection of $\mu^*$-measurable sets $\overline{\mathcal{S}}$ forms a ring. In fact, it forms a $\sigma$-ring. That is, given a countably infinite sequence ${E_i}_{i=1}^\infty$ of $\mu^*$-measurable sets, their union $E$ is also $\mu^*$-measurable. Even better if the $E_i$ are pairwise disjoint, then

$\displaystyle\mu^*(A\cap E)=\mu^*\left(A\cap\bigcup\limits_{i=1}^\infty E_i\right)=\sum\limits_{i=1}^\infty\mu^*(A\cap E_i)$

To see this, we start with a disjoint sequence and an equation we came up with yesterday:

\displaystyle\begin{aligned}\mu^*(A\cap(E_1\cup E_2))&=\mu^*(A\cap E_1\cap E_2)+\mu^*(A\cap E_1\cap E_2^c)+\mu^*(A\cap E_1^c\cap E_2)\\&=\mu^*(A\cap E_1)+\mu^*(A\cap E_2)\end{aligned}

We can keep going like this, adding in more and more of the $E_i$:

$\displaystyle\mu^*\left(A\cap\bigcup\limits_{i=1}^nE\right)=\sum\limits_{i=1}^n\mu^*(A\cap E_i)$

for every natural number $n$. This finite union $F_n=\bigcup_{i=1}^n E_i$ is $\mu^*$-measurable. This, along the fact that $F^c\subseteq E^c$, tells us that

\displaystyle\begin{aligned}\mu^*(A)&=\mu^*(A\cap F_n)+\mu^*(A\cap F^c)\\&\geq\sum\limits_{i=1}^n\mu^*(A\cap E_i)+\mu^*(A\cap E^c)\end{aligned}

This is true for every $n$, so we may pass to the limit and use the countable subadditivity of $\mu^*$

\displaystyle\begin{aligned}\mu^*(A)&\geq\sum\limits_{i=1}^\infty\mu^*(A\cap E_i)+\mu^*(A\cap E^c)\\&\geq\mu^*(A\cap E)+\mu^*(A\cap E^c)\end{aligned}

But this is enough to show that $E$ is $\mu^*$-measurable, and so $\overline{\mathcal{S}}$ is closed under countable disjoint unions. And this shows that $\overline{\mathcal{S}}$ is closed under countable unions in general, by our trick of replacing a sequence by a disjoint sequence with the same partial unions.

Since the previous inequalities must then actually be equalities, we see that we must have

$\displaystyle\sum\limits_{i=1}^\infty\mu^*(A\cap E_i)+\mu^*(A\cap E^c)\geq\mu^*(A\cap E)+\mu^*(A\cap E^c)$

It’s tempting to simply subtract $\mu^*(A\cap E^c)$ from both sides, but this might be an infinite quantity. Instead, we’ll simply replace $A$ with $A\cap E$, which has the same effect of giving us

$\displaystyle\sum\limits_{i=1}^\infty\mu^*(A\cap E_i)\geq\mu^*(A\cap E)$

as we claimed.

If we replace $A$ by $E$ in this equation, we find that — when restricted to the $\sigma$-ring $\overline{\mathcal{S}}$$\mu^*$ is actually countably additive. That is, if we define $\bar{\mu}(A)=\mu^*(A)$ for $A\in\overline{\mathcal{S}}$, then $\bar{\mu}$ is actually a measure. Even better, it’s a complete measure.

Indeed, if $E\subseteq F$ and $\bar{\mu}(F)=\mu^*(F)=0$, then $\mu^*(E)=0$ as well. We must show that $E$ is actually $\mu^*$-measurable, and so $\bar{\mu}(E)$ exists and equals zero. But we can easily see that for any $A\in\mathcal{H}$

$\displaystyle\mu^*(A)=\mu^*(E)+\mu^*(A)\geq\mu^*(A\cap E)+\mu^*(A\cap E^c)$

and this is enough to show that $E$ is $\mu^*$-measurable, and thus that $\bar{\mu}$ is complete.

March 30, 2010 Posted by | Analysis, Measure Theory | 5 Comments

## Sets Measurable by an Outer Measure I

An outer measure $\mu^*$ on a hereditary $\sigma$-ring $\mathcal{H}$ is nice and all, but it’s not what we really want, which is a measure. In particular, it’s subadditive rather than additive. We want to fix this by restricting to a nice collection of sets within $\mathcal{H}$.

Every set $E$ splits every other set into two pieces: the part that’s in $E$ and the part that’s not. What we want to focus on are the sets that split every other set additively. That is, sets $E\in\mathcal{H}$ so that for every set $A\in\mathcal{H}$ we have

$\displaystyle\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c)$

We call such sets “$\mu^*$-measurable”. Actually, to show that $E$ is $\mu^*$-measurable we just need to show that $\mu^*(A)\geq\mu^*(A\cap E)+\mu^*(A\cap E^c)$ for every $A\in\mathcal{H}$, because the opposite inequality follows from the subadditivity of $\mu^*$.

This condition seems sort of contrived at first, and there’s really not much to justify it at first besides the foregoing handwaving. But we will soon see that this definition turns out to be useful. For one thing, the collection $\overline{\mathcal{S}}\subseteq\mathcal{H}$ of $\mu^*$-measurable sets is a ring!

The proof of this fact is straightforward, but it feels like pulling a rabbit out of a hat, so follow closely. Given sets $\mu^*$-measurable sets $E$ and $F$, we need to show that their union $E\cup F$ and difference $E\setminus F=E\cap F^c$ are both $\mu^*$-measurable as well. Saying that $E$ is $\mu^*$-measurable means that for every $A\in\mathcal{H}$ we have

$\displaystyle\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c)$

Saying that $F$ is $\mu^*$-measurable means that for every $A\in\mathcal{H}$ we have

\displaystyle\begin{aligned}\mu^*(A\cap E)&=\mu^*(A\cap E\cap F)+\mu^*(A\cap E\cap F^c)\\\mu^*(A\cap E^c)&=\mu^*(A\cap E^c\cap F)+\mu^*(A\cap E^c\cap F^c)\end{aligned}

We can take each of these and plug them into the first equation to find the key equation

\displaystyle\begin{aligned}\mu^*(A)&=\mu^*(A\cap E\cap F)+\mu^*(A\cap E\cap F^c)\\&+\mu^*(A\cap E^c\cap F)+\mu^*(A\cap E^c\cap F^c)\end{aligned}

Now this key equation works for $A\cap(E\cup F)$ as well as $A$. We know that $(E\cup F)\cap E=E$ and $(E\cup F)\cap F=F$, but $(E\cup F)\cap E^c\cap F^c=\emptyset$. So, sticking $A\cap(E\cup F)$ into the key equation we find

\displaystyle\begin{aligned}\mu^*(A\cap(E\cup F))&=\mu^*(A\cap(E\cup F)\cap E\cap F)+\mu^*(A\cap(E\cup F)\cap E\cap F^c)\\&+\mu^*(A\cap(E\cup F)\cap E^c\cap F)+\mu^*(A\cap(E\cup F)\cap E^c\cap F^c)\\&=\mu^*(A\cap E\cap F)+\mu^*(A\cap E\cap F^c)+\mu^*(A\cap E^c\cap F)\end{aligned}

But the three terms on the right are the first three terms in the key equation. And so we can replace them and write

\displaystyle\begin{aligned}\mu^*(A)&=\mu^*(A\cap(E\cup F))+\mu^*(A\cap E^c\cap F^c)\\&=\mu^*(A\cap(E\cup F))+\mu^*(A\cap(E\cup F)^c)\end{aligned}

which establishes that $E\cup F$ is $\mu^*$-measurable! Behold, the rabbit!

Let’s see if we can do it again. This time, we take $A\cap(E\setminus F)^c=A\cap(E^c\cup F)$ and stick it into the key equation. We find

\displaystyle\begin{aligned}\mu^*(A\cap(E\setminus F)^c)&=\mu^*(A\cap(E^c\cup F)\cap E\cap F)+\mu^*(A\cap(E^c\cup F)\cap E\cap F^c)\\&+\mu^*(A\cap(E^c\cup F)\cap E^c\cap F)+\mu^*(A\cap(E^c\cup F)\cap E^c\cap F^c)\\&=\mu^*(A\cap E\cap F)+\mu^*(A\cap E^c\cap F)+\mu^*(A\cap E^c\cap F^c)\end{aligned}

Again we can find the three terms on the right of this equation on the right side of the key equation as well. Replacing them in the key equation, we find

\displaystyle\begin{aligned}\mu^*(A)&=\mu^*(A\cap(E\setminus F)^c)+\mu^*(A\cap(E\setminus F)^c)\end{aligned}

which establishes that $E\setminus F$ is $\mu^*$-measurable as well!

March 29, 2010 Posted by | Analysis, Measure Theory | 12 Comments

## Extending a Measure to an Outer Measure

Let $\mu$ be a measure in a ring (not necessarily an algebra) $\mathcal{R}\subseteq P(X)$, and let $\mathcal{H}=\mathcal{H}(\mathcal{R})$ be the hereditary $\sigma$-ring generated by $\mathcal{R}$. For every $E\in\mathcal{H}$, define

$\displaystyle\mu^*(E)=\inf\left\{\sum\limits_{i=1}^\infty\mu(E_i)\bigg\vert\{E_i\}\subseteq\mathcal{R}, E\subseteq\bigcup\limits_{i=1}^\infty E_i\right\}$

That is, $E\in\mathcal{H}$ can be covered by a countable collection of sets in $\mathcal{R}$. For every such cover, sum up the $\mu$-measures of all the sets in the cover, and define $\mu^*(E)$ to be the greatest lower bound of such sums. Then $\mu^*$ is an outer measure, which extends $\mu$ to all of $\mathcal{H}$. Further, if $\mu$ is $\sigma$-finite, then $\mu^*$ will be too. We call $\mu^*$ the outer measure “induced by” $\mu$.

First off, if $E\in\mathcal{R}$ itself, then we can cover it with itself and an infinite sequence of empty sets. That is, $E\subseteq E\cup\emptyset\cup\emptyset\cup\dots$. Thus we must have $\mu^*(E)\leq\mu(E)+\mu(0)+\dots=\mu(E)$. On the other hand, if $E\in\mathcal{R}$ is contained in the union of a sequence $\{E_i\}\subseteq\mathcal{E}$, then monotonicity tells us that $\mu(E)\leq\sum_i\mu(E_i)$, and thus $\mu(E)\leq\mu^*(E)$. That is, $\mu(E)$ must be equal to $\mu^*(E)$ for sets $E\in\mathcal{R}$; as a set function, $\mu^*$ indeed extends $\mu$. In particular, we find that $\mu^*(\emptyset)=\mu(\emptyset)=0$.

If $E$ and $F$ are sets in $\mathcal{H}$ with $E\subseteq F$ and $\{E_i\}$ is a sequence covering $F$, then it must cover $E$ as well. Thus $\mu^*(E)$ can be at most $\mu^*(F)$, and may be even smaller. This establishes that $\mu^*$ is monotonic.

We must show that $\mu^*$ is countably subadditive. Let $E\in\mathcal{H}$ and $\{E_i\}_{i=1}^\infty\subseteq\mathcal{H}$ be sets so that $E$ is contained in the union of the $E_i$. Let $\epsilon$ be an arbitrarily small positive number, and for each $i$ choose some sequence $\{E_{ij}\}_{j=1}^\infty\subseteq\mathcal{R}$ that covers $E_i$ such that

$\displaystyle\sum\limits_{j=1}^\infty\mu(E_{ij})\leq\mu^*(E_i)+\frac{\epsilon}{2^i}$

This is possible because the definition of $\mu^*$ tells us that we can find a covering sequence whose measure-sum exceeds $\mu^*(E_i)$ by an arbitrarily small amount. Then the collection of all the $E_{ij}$ constitute a countable collection of sets in $\mathcal{R}$ which together cover $E$. Thus we conclude that

$\displaystyle\mu^*(E)\leq\sum\limits_{i=1}^\infty\sum\limits_{j=1}^\infty\mu(E_{ij})\leq\sum\limits_{i=1}^\infty\left(\mu^*(E_i)+\frac{\epsilon}{2^i}\right)=\sum\limits_{i=1}^\infty\mu^*(E_i)+\epsilon$

Since $\epsilon$ was arbitrary, we conclude that

$\displaystyle\mu^*(E)\leq\sum\limits_{i=1}^\infty\mu^*(E_i)$

and so $\mu^*$ is countably subadditive.

Finally, if $E\in\mathcal{H}$, we can pick a cover $\{E_i\}_{i=1}^\infty\subseteq\mathcal{R}$. If $\mu$ is $\sigma$-finite, we can cover each of these sets by a sequence $\{E_{ij}\}_{j=1}^\infty\subseteq\mathcal{R}$ so that $\mu(E_{ij})<\infty$. The collection of all the $E_{ij}$ is then a countable cover of $E$ by sets of finite measure; the extension $\mu^*$ is thus $\sigma$-finite as well.

March 26, 2010 Posted by | Analysis, Measure Theory | 5 Comments

## Outer Measures

We’re going to want a modification of the notion of a measure. But before we introduce it, we have (of course) a few definitions.

First of all, a collection $\mathcal{E}\subseteq P(X)$ of sets is called “hereditary” if it includes all the subsets of each of its sets. That is, if $E\in\mathcal{E}$ and $F\subseteq E$, then $F\in\mathcal{E}$ as well. It’s not very useful to combine this with the definition of an algebra, because an algebra must contain $X$ itself; the only hereditary algebra is $P(X)$ itself. Instead, we define a “ring” of sets (or a $\sigma$-ring) to be closed under union (countable unions for $\sigma$-rings) and difference operations, but without the requirement that it contain $X$; complements and intersections are also not guaranteed, since we built these from differences using $X$ itself. Pretty much everything we’ve done so far with algebras can be done with rings, and hereditary $\sigma$-rings will be interesting objects of study.

Just like we found for algebras and monotone classes, the intersection of two hereditary collection is again hereditary. We can thus construct the “smallest” hereditary $\sigma$-ring containing a given collection $\mathcal{E}$, and we’ll call it $\mathcal{H}(\mathcal{E})$. In fact, it’s not hard to see that this is the collection of all sets which can be covered by a countable union of sets in $\mathcal{E}$; any $\sigma$-ring containing $\mathcal{E}$ must contain all such countable unions, and a hereditary collection must then contain all the subsets.

Now, an extended real-valued set function $\mu^*$ on a collection $\mathcal{E}$ is called “subadditive” whenever $E$, $F$, and their union $E\cup F$ are in $\mathcal{E}$, we have the inequality

$\displaystyle\mu^*(E\cup F)\leq\mu^*(E)+\mu^*(F)$

It’s called “finitely subadditive” if for every finite collection $\{E_1,\dots,E_n\}\subseteq\mathcal{E}$ whose union is also contained in $\mathcal{E}$ we have the inequality

$\displaystyle\mu^*\left(\bigcup\limits_{i=1}^nE_i\right)\leq\sum\limits_{i=1}^n\mu^*(E_i)$

and “countably subadditive” if for every sequence $\{E_i\}_{i=1}^\infty$ of sets in $\mathcal{E}$ whose union is also in $\mathcal{E}$, we have

$\displaystyle\mu^*\left(\bigcup\limits_{i=1}^\infty E_i\right)\leq\sum\limits_{i=1}^\infty\mu^*(E_i)$

Note that these differ from additivity conditions in two ways: we only ask for an inequality to hold, and we don’t require the unions to be disjoint.

Finally, we can define an “outer measure” to be an extended real-valued, non-negative, monotone, and countably subadditive set function $\mu^*$, defined on a hereditary $\sigma$-ring $\mathcal{H}$, and such that $\mu^*(\emptyset)=0$. Just as for a measure, we say that $\mu^*$ is “finite” or “$\sigma$-finite” if every set has finite or $\sigma$-finite outer measure.

March 25, 2010 Posted by | Analysis, Measure Theory | 12 Comments

## Measure as Metric

It turns out that a measure turns its domain into a sort of metric space, measuring the “distance” between two sets. So, let’s say $\mu$ is a measure on an algebra $\mathcal{A}$, and see how this works.

First, we need to define the “symmetric difference” of two sets. This is the collection of points in one or the other set, but not in both:

$\displaystyle A\Delta B = (A\setminus B)\cup(B\setminus A)$

So now we define an extended real-valued function $d:\mathcal{A}\times\mathcal{A}\to\overline{\mathbb{R}}$ by

$\displaystyle d(A,B)=\mu(A\Delta B)$

I say that this has almost all the properties of a metric function. First of all, it’s clearly symmetric and nonnegative, so that’s two of the four right there. It also satisfies the triangle inequality. That is, for any three sets $A$, $B$, and $C$ in $\mathcal{A}$, we have the inequality

$\displaystyle d(A,B)\leq d(A,C)+d(C,B)$

Indeed, points in $A\setminus B$ are either in $C$ or not. If not, then they’re in $A\setminus C$, while if they are they’re in $C\setminus B$. Similarly, points in $B\setminus A$ are either in $B\setminus C$ or $C\setminus A$. That is, the symmetric difference $A\Delta B$ is contained in the union of the symmetric difference $A\setminus C$ and the symmetric difference $C\setminus B$. And so monotonicity tells us that

$\displaystyle\mu(A\Delta B)\leq\mu((A\Delta C)\cup(C\Delta B))\leq\mu(A\Delta C)+\mu(C\Delta B)$

establishing the triangle inequality.

What’s missing is the assertion that $d(A,B)=0$ if and only if $A=B$. But there may be plenty of sets with measure zero, and any one of them could arise as a symmetric difference; as written, our function $d$ is not a metric. But we can fix this by changing the domain.

Let’s define a relation: $A\sim B$ if and only if $\mu(A\Delta B)=0$. This is clearly reflexive and symmetric, and the triangle inequality above shows that it’s transitive. Thus $\sim$ is an equivalence relation, and we can pass to the collection of equivalence classes. That is, we consider two sets $A$ and $B$ to be “the same” if $A\sim B$.

This trick will handle the obstruction to $d$ being a metric, but only if we can show that $d$ gives a well-defined function on these equivalence classes. That is, if $A\sim A'$ and $B\sim B'$, then $d(A,B)=d(A',B')$. But $A\sim A'$ means $d(A,A')=0$, and similarly for $B$. Thus we find

$\displaystyle d(A,B)\leq d(A,A')+d(A',B')+d(B',B)=d(A',B')$

and, similarly

$\displaystyle d(A',B')\leq d(A',A)+d(A,B)+d(B,B')=d(A,B)$

and so the two are equal. We define the distance between two $\sim$-equivalence classes by picking a representative of each one and calculating $d$ between them.

This relation $\sim$ turns out to be extremely useful. That is, as we go forward we will often find things simpler if we consider two sets to be “the same” if they differ by a set of measure zero, or by a subset of such a set. We will call subsets of sets of measure zero “negligible”, since we can neglect things that only happen on such a set.

March 24, 2010 Posted by | Analysis, Measure Theory | 10 Comments

## 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

## Monotonicity of Measures

First, we make a couple of definitions. A set function $\mu$ on a collection of sets $\mathcal{E}$ is “monotone” if whenever we have $E_1$ and $E_2$ in $\mathcal{E}$ with $E_1\subseteq E_2$, then $\mu(E_1)\leq\mu(E_2)$. We say that $\mu$ is “subtractive” if whenever further $E_2\setminus E_1\in\mathcal{E}$ and $\lvert\mu(E_1)\rvert<\infty$, then $\mu(E_2\setminus E_1)=\mu(E_2)-\mu(E_1)$.

Now I say that any measure on an algebra $\mathcal{A}$ is both monotone and subtractive. Indeed, since $\mathcal{A}$ is an algebra, then $E_2\setminus E_1$ is guaranteed to be in $\mathcal{A}$ as well, and it’s disjoint from $E_1$. And so we see that

$\displaystyle\mu(E_2)=\mu(E_1\uplus(E_2\setminus E_1))=\mu(E_1)+\mu(E_2\setminus E_1)$

Since $\mu$ is non-negative, we must have $\mu(E_2)\geq\mu(E_1)$, and so $\mu$ is monotone. And if $\mu(E_1)$ is finite, then we can subtract it from both sides of this equation to show that $\mu$ is subtractive as well.

Next, say $\mu$ is a measure on an algebra $\mathcal{A}$. If we have a set $A\in\mathcal{A}$ and a finite or (countably) infinite sequence of sets $\{A_i\}$ so that $A\subseteq\bigcup_iA_i$, then we have the inequality $\mu(A)\leq\sum_i\mu(A_i)$. To show this, we’ll invoke a very useful trick: if $\{F_i\}$ is a sequence of sets in an algebra $A$, then there is a pairwise disjoint sequence $\{G_i\}$ of sets so that each $G_i\subseteq F_i$ and $\bigcup_iG_i=\bigcup_iF_i$. Indeed, we define

$\displaystyle G_i=F_i\setminus\bigcup\limits_{1\leq j

That is, $G_1$ is the same as $F_1$, and after that point each $G_i$ is everything in $F_i$ that hasn’t been covered already.

So we start with $B_i=A\cap A_i$, and come up with a new sequence $\{C_i\}$. The (disjoint) union of the $C_i$ is, $A$ like the union of the $B_i$. Then, by the countable additivity of $\mu$, we have $\mu(A)=\sum_i\mu(C_i)$. But the monotonicity says that $\mu(C_i)\leq\mu(B_i)$, and also that $\mu(B_i)\leq\mu(A_i)$, and so $\mu(A)\leq\sum_i\mu(A_i)$.

On the other hand, if $\{A_i\}$ is a pairwise disjoint and we have $\bigcup_iA_i\subseteq A$, then $\sum_i\mu(A_i)\leq\mu(A)$. Indeed, if $\{A_i\}$ is a finite sequence, then the union $\bigcup_iA_i$ is in $\mathcal{A}$. Monotonicity and finite additivity then tell us that

$\displaystyle\sum\limits_i\mu(A_i)=\mu\left(\bigcup\limits_iA_i\right)\leq\mu(A)$

However if $\{A_i\}$ is an infinite sequence, then what we just said applies to any finite subsequence. Then the sum $\sum_{i=1}^\infty\mu(A_i)$ is the limit of the sums $\sum_{i=1}^n\mu(A_i)$. Each of these is less than or equal to $\mu(A)$, and so their limit must be as well.

March 22, 2010 Posted by | Analysis, Measure Theory | 8 Comments

## Measures

From this point in, I will define a “set function” as a function $\mu$ whose domain is some collection of subsets $\mathcal{E}\subseteq P(X)$. It’s important to note here that $\mu$ is not defined on points of the set $X$, but on subsets of $X$. For some reason, a lot of people find that confusing at first.

We’re primarily concerned with set functions which take their values in the “extended real numbers” $\overline{\mathbb{R}}$. That is, the value of $\mu(E)$ is either a real number, or $+\infty$, or $-\infty$, with the latter two being greater than all real numbers and less than all real numbers, respectively.

We say that such a set function $\mu:\mathcal{E}\to\overline{\mathbb{R}}$ is “additive” if whenever we have disjoint sets $E_1$ and $E_2$ in $\mathcal{E}$ with disjoint union $E_1\uplus E_2$ also in $\mathcal{E}$, then we have

$\displaystyle\mu(E_1\uplus E_2)=\mu(E_1)+\mu(E_2)$

Similarly, we say that $\mu$ is finitely additive if for every finite, pairwise disjoint collection $\{E_1,\dots,E_2\}\subseteq\mathcal{E}$ whose union is also in $\mathcal{E}$ we have

$\displaystyle\mu\left(\biguplus\limits_{i=1}^nE_i\right)=\sum\limits_{i=1}^n\mu(E_i)$

And we say that $\mu$ is countably additive of for every pairwise-disjoint sequence $\{E_i\}_{i=1}^\infty$ of sets in $\mathcal{E}$ whose union is also in $\mathcal{E}$, we have

$\displaystyle\mu\left(\biguplus\limits_{i=1}^\infty E_i\right)=\sum\limits_{i=1}^\infty\mu(E_i)$

Now we can define a “measure” as an extended real-valued, non-negative, countably additive set function $\mu$ defined on an algebra $\mathcal{A}$, and satisfying $\mu(\emptyset)=0$. With this last assumption, we can show that a measure is also finitely additive. Indeed, given a collection $\{E_1,\dots,E_n\}$, just define $E_i=\emptyset$ for $i>n$ to get a sequence. Then we find

$\displaystyle\mu\left(\biguplus\limits_{i=1}^nE_i\right)=\mu\left(\biguplus\limits_{i=1}^\infty E_i\right)=\sum\limits_{i=1}^\infty\mu(E_i)=\sum\limits_{i=1}^n\mu(E_i)$

If $\mu$ is a measure on $\mathcal{A}$, we say a set $A\in\mathcal{A}$ has finite measure if $\mu(A)<\infty$. We say that $A$ has “$\sigma$-finite” measure if there is a sequence of sets $\{A_i\}_{i=1}^\infty$ of finite measure ($\mu(A_i)<\infty$) so that $A\subseteq\bigcup_{i=1}^\infty A_i$. If every set in $A$ has finite (or $\sigma$-finite) measure, we say that $\mu$ is finite (or $\sigma$-finite) on $A$.

Finally, we say that a measure is “complete” if for every set $A$ of measure zero, $\mathcal{A}$ also contains all subsets of $A$. That is, if $A\in\mathcal{A}$, $\mu(A)=0$, and $B\subseteq A$, then $B\in\mathcal{A}$. At first, this might seem to be more a condition on the algebra $\mathcal{A}$ than on the measure $\mu$, but it really isn’t. It says that to be complete, a measure can only assign ${0}$ to a set if all of its subsets are also in $\mathcal{A}$.

March 19, 2010 Posted by | Analysis, Measure Theory | 17 Comments

## Monotone Classes

Now we want to move from algebras of sets closer to $\sigma$-algebras by defining a new structure: a “monotone class”. This is a collection $\mathcal{M}\subseteq P(Z)$ so that if we have an increasing sequence of sets

$\displaystyle E_1\subseteq E_2\subseteq E_3\subseteq\dots$

with $E_i\in\mathcal{M}$, then the “limit” of this sequence is in $\mathcal{M}$. Similarly, if we have a decreasing sequence of sets

$\displaystyle E_1\supseteq E_2\supseteq E_3\supseteq\dots$

with $E_i\in\mathcal{M}$, then the “limit” of this sequence is in $\mathcal{M}$. In each case, we can construe this limit more simply. In the case of an increasing sequence, each term $E_i$ contains all the terms before it, and is thus the same as the union of the whole sequence up to that point. Thus we can define

$\displaystyle\lim\limits_{n\to\infty}E_n=\lim\limits_{n\to\infty}\bigcup\limits_{i=1}^nE_i=\bigcup\limits_{i=1}^\infty E_i$

Similarly, for a decreasing sequence we can define

$\displaystyle\lim\limits_{n\to\infty}E_n=\lim\limits_{n\to\infty}\bigcap\limits_{i=1}^nE_i=\bigcap\limits_{i=1}^\infty E_i$

Now, if $\mathcal{A}\subseteq P(X)$ is an algebra of subsets of $X$, and $\mathcal{M}$ is a monotone class containing $\mathcal{A}$, then $\mathcal{M}$ also contains $\mathcal{S}(\mathcal{A})$, the smallest $\sigma$-algebra containing $\mathcal{A}$.

As an aside, what do we mean by “smallest”? Well, it’s not hard to see that the intersection of a collection of $\sigma$ algebras is itself a $\sigma$-algebra, just like we did way back when we showed that we had lower bounds in the lattice of ideals. So take all the $\sigma$-algebras containing $\mathcal{A}$ — there is at least one, because $P(X)$ itself is one — and define $\mathcal{S}(\mathcal{A})$ to be the intersection of all of them. This is a $\sigma$-algebra contained in all the others, thus “smallest”.

Similarly, there is a smallest monotone class containing $\mathcal{A}$; we will assume that $\mathcal{M}$ is this one without loss of generality, since if it contains $\mathcal{S}(\mathcal{A})$ then all the others do as well. We will show that this $\mathcal{M}$ is actually a $\sigma$-algebra, and then it must contain $\mathcal{S}(\mathcal{A})$!

Given $A\in\mathcal{A}$, define $\mathcal{M}_A$ to be the collection of $B\in\mathcal{M}$ so that $A\cup B\in\mathcal{M}$. This is itself a monotone class, and it contains $\mathcal{A}$, and so it must contain $\mathcal{M}$. But it’s defined as being contained in $\mathcal{M}$, and so we must have $\mathcal{M}_A=\mathcal{M}$. Thus $A\cup B\in\mathcal{M}$ as long as $A\in\mathcal{A}$ and $B\in\mathcal{M}$.

Now take a $B\in\mathcal{M}$ and define $\mathcal{M}_B$ to be the collection of $D\in\mathcal{M}$ so that $D\cup B\in\mathcal{M}$. What we just showed is that $\mathcal{A}\subseteq\mathcal{M}_B$, and $\mathcal{M}_B$ is a monotone class. And so $\mathcal{M}_B-\mathcal{M}$, and $D\cup B\in\mathcal{M}$ whenever both $D$ and $B$ are in $\mathcal{M}$.

We can repeat a similar argument to the previous two paragraphs to show that $D\setminus B\in\mathcal{M}$ for $D,B\in\mathcal{M}$. Start by defining $\mathcal{M}_A$ as the collection of $B\in\mathcal{M}$ so that both $B\setminus A$ and $A\setminus B$ are in $\mathcal{M}$.

So $\mathcal{M}$ is both a monotone class and an algebra. We need to show that it’s a $\sigma$-algebra by showing that it’s closed under countable unions. But if $E_i\in\mathcal{M}$ then $F_n=\cup_{i=1}^nE_i$ is a monotone increasing sequence of set in $\mathcal{M}$, since they’re all finite unions. The countable union is the limit of this sequence, which $\mathcal{M}$ contains by virtue of being a monotone class.

And so $\mathcal{M}$ is a $\sigma$-algebra containing $\mathcal{A}$, and so it contains $\mathcal{S}(\mathcal{A})$.

March 18, 2010 Posted by | Analysis, Measure Theory | 4 Comments