The Unapologetic Mathematician

Measurable (Extended) Real-Valued Functions

For a while, we’ll mostly be interested in real-valued functions with Lebesgue measure on the real line, and ultimately in using measure to give us a new and more general version of integration. When we couple this with our slightly weakened definition of a measurable space, this necessitates a slight tweak to our definition of a measurable function.

Given a measurable space $(X,\mathcal{S})$ and a function $f:X\to\mathbb{R}$, we define the set $N(f)$ as the set of points $x\in X$ such that $f(x)\neq0$. We will say that the real-valued function $f$ is measurable if $N(f)\cap f^{-1}(M)$ is a measurable subset of $X$ for every Borel set $M\in\mathcal{B}$ of the real line. We have to treat ${0}$ specially because when we deal with integration, ${0}$ is special — it’s the additive identity of the real numbers.

The entire real line $\mathbb{R}$ is a Borel set, and $f^{-1}(\mathbb{R})=X$. Thus we find that $N(f)$ must be a measurable subset of $X$. If $E$ is another measurable subset of $X$, then we observe

$\displaystyle E\cap f^{-1}(M)=((E\cap N(f))\cap f^{-1}(M))\cup((E\setminus N(f))\cap f^{-1}(M))$

The second term on the right is either empty or is equal to $E\setminus N(f)$. And so it’s clear that $E\cap f^{-1}(M)$ is measurable. We say that the function $f$ is “measurable on $E$” if $E\cap f^{-1}(M)$ is measurable for every Borel set $M$, and so we have shown that a measurable function is measurable on every measurable set.

In particular, if $X$ is itself measurable (as it often is), then a real-valued function is measurable if and only if $f^{-1}(M)$ is measurable for every Borel set $M\in\mathcal{B}$. And so in this (common) case, we get back our original definition of a measurable function $f:(X,\mathcal{S})\to(\mathbb{R},\mathcal{B})$.

The concept of measurability depends on the $\sigma$-ring $\mathcal{S}$, and we sometimes have more than one $\sigma$-ring floating around. In such a case, we say that a function is measurable with respect to $\mathcal{S}$. In particular, we will often be interested in the case $X=\mathbb{R}$, equipped with either the $\sigma$-algebra of Borel sets $\mathcal{B}$ or that of Lebesgue measurable sets $\overline{\mathcal{B}}$. A measurable function $f:(\mathbb{R},\mathcal{B})\to(\mathbb{R},\mathcal{B})$ will be called “Borel measurable”, while a measurable function $f:(\mathbb{R},\overline{\mathcal{B}})\to(\mathbb{R},\mathcal{B})$ will be called “Lebesgue measurable”.

On the other hand, we should again emphasize that the definition of measurability does not depend on any particular measure $\mu$.

We will also sometimes want to talk about measurable functions taking value in the extended reals. We take the convention that the one-point sets $\{\infty\}$ and $\{-\infty\}$ are Borel sets; we add the requirement that a real-valued function also have $f^{-1}(\{\infty\})$ and $f^{-1}(\{-\infty\})$ both be measurable to the condition for $f$ to be measurable. However, for this extended concept of Borel sets, we can no longer generate the class of Borel sets by semiclosed intervals.

April 30, 2010 Posted by | Analysis, Measure Theory | 6 Comments

Measurable Subspaces III

To recap: we’ve got a measure space $(X,\mathcal{S},\mu)$ and we’re talking about what structure we get on a subset $X_0$. If $X_0$ is measurable — if $X_0\in\mathcal{S}$ — then we can set $\mathcal{S}_0$=$\mathcal{S}\cap X_0=\{M\cap X_0\vert M\in\mathcal{S}\}$. Since each of these subsets $M\cap X_0$ is itself measurable as a subset of $X$, we can just define $\mu_0(M\cap X_0)=\mu(M\cap X_0)$. On the other hand, if $X_0$ is nonmeasurable but thick we can use the same definition for $\mathcal{S}_0$. This time, though, the subsets $M\cap X_0$ may not themselves be in $\mathcal{S}$, and so we can’t do the same thing. We saw, though, that we can define $\mu_0(M\cap X_0)=\mu(M)$.

So what if $X_0$ is neither measurable nor thick? It turns out that if we want to use this latter method of defining $\mu_0$, $X_0$ must be thick! In particular, in order to prove that $\mu_0$ is well-defined we had to show that if $M_1$ and $M_2$ are two measurable subsets of $X$ with $M_1\cap X_0=M_2\cap X_0$, then $\mu(M_1)=\mu(M_2)$. I say that if $M_1\cap X_0=M_2\cap X_0$ implies $\mu(M_1)=\mu(M_2)$ for any two measurable sets $M_1$ and $M_2$, then $X_0$ must be thick.

To see this, first take any measurable set $E$ and pick another one $F\subseteq E\setminus X_0$. Then we see that $(E\setminus F)\cap X_0=E\cap X_0$, and so our hypothesis tells us that $\mu(E)=\mu(E\setminus F)$. Since $F\subseteq E\setminus X_0\subseteq E$, the subtractivity of $\mu$ tells us that $\mu(E)=\mu(E\setminus F)=\mu(E)-\mu(F)$, and we conclude that $\mu(F)=0$. That is, every measurable set $F$ that fits into $E\setminus X_0$ must have measure zero, and thus $\mu_*(E\setminus X_0)$ — as the supremum of the measures of all these sets — must be zero as well.

And so we see that in order for this $\mu_0$ to be unambiguously defined, we must require that $X_0$ be thick.

April 29, 2010

Measurable Subspaces II

Last time we discussed how to define a measurable subspace $(X_0,\mathcal{S}_0)$ of a measurable space $(X,\mathcal{S})$ in the easy case when $X_0$ is itself a measurable subset of $X$: $X_0\in\mathcal{S}$.

But what if $X_0$ isn’t measurable as a subset of $X$? To get at this question, we introduce the notion of a “thick” subset. We say that a subset $X_0$ of a measure space $(X,\mathcal{S},\mu)$ is thick if $\mu_*(E\setminus X_0)=0$ for all measurable $E\in\mathcal{S}$. If $X$ is itself measurable (as it often is), this condition reduces to asking that $\mu_*(X\setminus X_0)=0$. If, further, $\mu(X)<\infty$, then we ask that $\mu^*(X_0)=\mu(X)$. As an example, the maximally nonmeasurable set we constructed is thick.

Now I say that if $X_0$ is a thick subset of a measure space $(X,\mathcal{S},\mu)$, if $\mathcal{S}_0=\{M\cap X_0\vert M\in\mathcal{S}\}$ consists of all intersections of $X_0$ with measurable subsets of $X$, and if $\mu_0$ is defined by $\mu_0(M\cap X_0)=\mu(M)$, then $(X_0,\mathcal{S}_0,\mu_0)$ is a measure space. This definition of $\mu_0$ is unambiguous, since if $M_1$ and $M_2$ are two measurable subsets of $X$ with $M_1\cap X_0=M_2\cap X_0$, then $(M_1\Delta M_2)\cap X_0=\emptyset$. The thickness of $X_0$ implies that $\mu_*((M_1\Delta M_2)\cap X_0^c)=0$, and we know that

$\displaystyle\mu_*((M_1\Delta M_2)\cap X_0^c)+\mu^*((M_1\Delta M_2)\cap X_0)=\mu(M_1\Delta M_2)$

Since $(M_1\Delta M_2)\cap X_0=\emptyset$, the second term must be zero, and so $\mu(M_1\Delta M_2)=0$. Therefore, $\mu(M_1)=\mu(M_2)$, and $\mu_0$ is indeed unambiguously defined.

Now given a pairwise disjoint sequence $\{F_n\}$ of sets in $\mathcal{S}_0$, define $\{E_n\}\subseteq\mathcal{S}$ to be measurable sets so that $F_n=E_n\cap X_0$. If we define

$\displaystyle\tilde{E}_n=E_n\setminus\bigcup\limits_{1\leq i

then we find

\displaystyle\begin{aligned}(\tilde{E}_n\Delta E_n)\cap X_0&=\left(\left(E_n\setminus\bigcup\limits_{1\leq i

and so $\mu(\tilde{E}_n\Delta E_n)=0$. Therefore

\displaystyle\begin{aligned}\sum\limits_{n=1}^\infty\mu_0(F_n)&=\sum\limits_{n=1}^\infty\mu(E_n)\\&=\sum\limits_{i=1}^\infty\mu(\tilde{E}_n)\\&=\mu\left(\bigcup\limits_{i=1}^\infty\tilde{E}_n\right)\\&=\mu\left(\bigcup\limits_{i=1}^\infty E_n\right)\\&=\mu_0\left(\bigcup\limits_{i=1}^\infty F_n\right)\end{aligned}

which shows that $\mu_0$ is indeed a measure.

April 28, 2010 Posted by | Analysis, Measure Theory | 3 Comments

Measurable Subspaces I

WordPress seems to have cleaned up its mess for now, so I’ll try to catch up.

When we’re considering the category of measurable spaces it’s a natural question to ask whether a subset $X_0\subseteq X$ of a measurable space $(X,\mathcal{S})$ is itself a measurable space in a natural way, and if this constitutes a subobject in the category. Unfortunately, unlike we saw with topological spaces, it’s not always possible to do this with measurable spaces. But let’s see what we can say.

Every subset comes with an inclusion function $\iota:X_0\hookrightarrow X$. If this is a measurable function, then it’s clearly a monomorphism; our question comes down to whether the inclusion is measurable in the first place. And so — as we did with topological spaces — we consider the preimage $\iota^{-1}(M)$ of a measurable subset $M\subseteq X$. That is, what points $x\in X_0$ satisfy $\iota(x)\in M$? Clearly, these are the points in the intersection $X_0\cap M$. And so for $\iota$ to be measurable, we must have $X_0\cap M$ be measurable as a subset of $X_0$.

An easy way for this to happen is for $X_0$ itself to be measurable as a subset of $X$. That is, if $X_0\in\mathcal{S}$, then for any measurable $M\in\mathcal{S}$, we have $X_0\cap M\in\mathcal{S}$. And so we can define $\mathcal{S}_0$ to be the collection of all measurable subsets of $X$ that happen to fall within $X_0$. That is, $M\in\mathcal{S}_0$ if and only if $M\in\mathcal{S}$ and $M\subseteq X_0$. If $X$ is a measure space, with measure $\mu$, then we can define a measure $\mu_0$ on $\mathcal{S}_0$ by setting $\mu_0(M)=\mu(M)$. This clearly satisfies the definition of a measure.

Conversely, if $(X_0,\mathcal{S}_0,\mu_0)$ is a measure space and $X_0\subseteq X$, we can make $X$ into a measure space $(X,\mathcal{S},\mu)$! A subset $M\subseteq X$ is in $\mathcal{S}$ if and only if $M\cap X_0\in\mathcal{S}_0$, and we define $\mu(M)=\mu_0(M\cap X_0)$ for such a subset $M$.

As a variation, if we already have a measurable space $(X,\mathcal{S})$ we can restrict it to the measurable subspace $(X_0,\mathcal{S}_0)$. If we then define a measure $\mu_0$ on $(X_0,\mathcal{S}_0)$, we can extend this measure to a measure $\mu$ on $(X,\mathcal{S})$ by the same definition: $\mu(M)=\mu_0(M\cap X_0)$, even though this $\mathcal{S}$ is not the same one as in the previous paragraph.

April 28, 2010 Posted by | Analysis, Measure Theory | 3 Comments

Things break down

WordPress seems to have messed with $\LaTeX$ again, and fouled it all up. Dozens of perfectly well-formed expressions are throwing errors, including $latex \sigma$. Since writing lowercase sigmas is pretty much essential for the current topics, I’m just not going to write until they fix their mess.

And if anyone from WordPress reads this: one of the major I encouraged people to start math, physics, and computer science oriented weblogs on WordPress’ platform is exactly its support for LaTeX. It really annoys me to no end that you keep screwing with it and breaking it in pretty severe ways.

April 28, 2010 Posted by | Uncategorized | 2 Comments

Measurable Spaces, Measure Spaces, and Measurable Functions

We’ve spent a fair amount of time discussing rings and $\sigma$-rings of sets, and measures as functions on such collections. Now we start considering how these sorts of constructions relate to each other.

A “measurable space” is some set $X$ and a choice of a $\sigma$-ring $\mathcal{S}$ of subsets of $X$. We call the members of $\mathcal{S}$ the “measurable sets” of the measurable space. This is not to insinuate that $\mathcal{S}$ is the collection of sets measurable by some outer measure $\mu^*$, nor even that we can define a nontrivial measure on $\mathcal{S}$ in the first place. Normally we just call the measurable space by the same name as the underlying set $X$ and omit explicit mention of $\mathcal{S}$.

Since it would be sort of silly to have points that can’t be discussed, we add the assumption that every point of $X$ is in some measurable set. Commonly, it’s the case that $X$ itself is measurable — $X\in\mathcal{S}$ — but we won’t actually require that $\mathcal{S}$ be a $\sigma$-algebra.

A “measure space” is a measurable space along with a choice of a measure $\mu$ on the $\sigma$-ring $\mathcal{S}$. As before, we will usually call a measure space by the same name as the underlying set $X$ and omit explicit mention of the $\sigma$-ring and measure. Measure spaces inherit adjectives from their measures; a measure space is called finite, or $\sigma$-finite, or complete if its measure $\mu$ is finite, $\sigma$-finite, or complete, respectively.

Given a measure space $(X,\mathcal{S},\mu)$, we will routinely use without comment the associated outer measure $\mu^*$ and inner measure $\mu_*$ on the hereditary $\sigma$-ring $\mathcal{H}(\mathcal{S})$.

As an underlying set equipped with a particular collection of “special” subsets, a measurable space should remind us of a topological space, and like topological spaces they form a category. Remember that our original definition of a continuous function: given topological spaces $(X_1,\mathcal{T}_1)$ and $(X_2,\mathcal{T}_2)$, a function $f:X_1\to X_2$ is continuous if the preimage of any open set is open — for any $O\in\mathcal{T}_2$ we have $f^{-1}(O)\in\mathcal{T}_1$.

We define a “measurable function” similarly: given measurable spaces $(X_1,\mathcal{S}_1)$ and $(X_2,\mathcal{S}_2)$, a function $f:X_1\to X_2$ is measurable if the preimage of any measurable set is measurable — for any $S\in\mathcal{S}_2$ we have $f^{-1}(S)\in\mathcal{S}_1$. It’s straightforward to verify that the collection of measurable spaces and measurable functions forms a category. We will set this category in its full generality aside for the moment, as is the usual practice in measure theory, but we will refer to it if appropriate to illuminate a point.

Before I close, though, I’d like to put out a question that I don’t know the answer to, and which some friends haven’t really been able to answer when I mused about it in front of them. When we dealt with topology, we were able to recast the basic foundations in terms of nets. That is, a function is continuous if and only if it “preserves limits of convergent nets” — if it sends any convergent net in the domain to another convergent net in the range, and the action of the function commutes with passage to the limit. I like this because the idea of “preserving” some structure (albeit an infinite and often-messy one) feels more natural and algebraic that the idea of inspecting preimages of open sets. And so I put the question out to the audience: what is “preserved” by a measurable function, in the same way that continuous functions preserve limits of convergent nets?

April 26, 2010 Posted by | Analysis, Measure Theory | 14 Comments

Non-Lebesgue Measurable Sets

I need to make up for missing a post earlier this week…

The most important observation about the fact that Lebesgue measurable sets might not be all of $P(X)$ is sort of tautological: it means that there may be subsets of the real line which are not Lebesgue measurable. That is, sets for which it is impossible to give a sense of “how much space they take up”, in a way compatible with the length of an interval.

And we can show that such sets do, in fact, exist. At least, we can build them if we have use of the axiom of choice. This might seem like a reason not to use the axiom of choice, but remember that Zorn’s lemma — which is equivalent to the axiom of choice — was essential when we needed to show that every vector space has a basis, or Tychonoff’s theorem, or that exact sequences of vector spaces split. So it’s sort of a mixed bag. In practice, most working mathematicians seem to be willing to accept the existence of non-Lebesgue measurable sets in order to gain the above benefits.

So, first a lemma: if $\xi$ is irrational, then the set $A$ of all numbers of the form $n+m\xi$ with $n$ and $m$ any integers is dense in the real line. That is, every open interval $U$ contains at least one point of $A$. The same is true for the set $B$ where we restrict $n$ to be even, and for the set $C$ where we restrict $n$ to be odd. Note $A$ (and, incidentally, $B$) is actually a subgroup of the additive group of real numbers.

For every integer $i$ there is some unique integer $n_i$ so that $0\leq n_i+i\xi<1$; we will write $x_i=n_i+i\xi$. If $U$ is an open interval, there is some positive integer $k$ with $\mu(U)>\frac{1}{k}$. Picking out the first $k+1$ numbers $x_1,\dots,x_{k+1}$, there must be some pair $x_i$ and $x_j$ with $\lvert x_i-x_j\rvert<\frac{1}{k}$ (or else they wouldn’t all fit in the interval $\left[0,1\right)$). But then some multiple of $x_i-x_j$ must land within $U$, as we asserted. For $B$, we can do the same using the interval $\left[0,2\right)$, and for $C$ we can use the fact that $C=B+1$.

Now, I say that there exists at least one set $E_0$ which is not Lebesgue measurable. To show this, we consider the quotient group $\mathbb{R}/A$. That is, we use an equivalence relation $x\sim y$ if $x-y\in A$. This divides up the real numbers into a disjoint union of equivalence classes under this relation, and the axiom of choice allows us to build a set $E_0$ by picking exactly one point from each equivalence class. This is the set we will show is not Lebesgue measurable.

Suppose $F$ is a Borel set contained in $E_0$. The difference set $D(F)$ contains no point of $A$, since if this happened we’d have two points in $E_0$ picked from the same $\sim$-equivalence class. But we just saw that any open interval contains a point of $A$, and so our result from last time shows that $F$ must have outer measure zero — if it had positive outer measure then $D(F)$ would contain an open interval. And so if $E_0$ is Lebesgue measurable then its Lebesgue measure must be zero.

Now if $a_1$ and $a_2$ are distinct elements of $A$, then $E_0+a_1$ and $E_0+a_2$ must be disjoint. As we let $a$ range over the countable number of values in $A$, the sets $E_0+a$ then form a countable disjoint cover of $\mathbb{R}$. But each of the $E_0+a$ is just a translation of $E_0$, and so each one must have the same measure. And then since Lebesgue measure is countably additive, we must have

$\displaystyle\bar{\mu}\left(\mathbb{R}\right)=\bar{\mu}\left(\bigcup\limits_{a\in A}E_0+a\right)=\sum\limits_{a\in A}\bar{\mu}(E_0+a)=\sum\limits_{a\in A}0=0$

But this is clearly nonsense.

We can do even better, actually, in our efforts to find bizarre sets. There exists a subset $M\subseteq\mathbb{R}$ so that for every Lebesgue measurable set $E$ we have both

$\displaystyle\mu_*(M\cap E)=0$
$\displaystyle\mu^*(M\cap E)=\bar{\mu}(E)$

That is, no matter what Lebesgue measurable set we pick, its intersection with $M$ is so weird that no set of positive Lebesgue measure can fit inside it, and yet $E$ itself is the smallest Lebesgue measurable sets that can contain it.

To find this set, write $A=B\cup C$ from our lemma and take $E_0$ to the set we just constructed. Define $M=E_0+B$ — the set of sums of points in $E_0$ and points in $B$. If $F$ is a Borel set contained in $M$, then $D(F)$ can’t contain any point of $C$ (using a similar argument to that from earlier). And so we must have $\mu_*(M)=0$.

On the other hand, we just saw that $E_0+A=\mathbb{R}$, and thus

$\displaystyle M^c=E_0+C=E_0+(B+1)=M+1$

And so $\mu_*(M^c)=0$ as well. If $E$ is any Lebesgue measurable set, the monotonicity of $\mu_*$ gives us $\mu_*(M\cap E)=\mu_*(M^c\cap E)=0$. And then an earlier result tells us that

$\displaystyle\bar{\mu}(E)=\mu_*(M^c\cap E)+\mu^*((M^c)^c\cap E)=0+\mu^*(M\cap E)=\mu^*(M\cap E)$

April 24, 2010 Posted by | Analysis, Measure Theory | 7 Comments

Lebesgue Measurable Sets

The attentive reader will note that in our study of Lebesgue measure we’ve defined it on some complete $\sigma$-algebra $\overline{\mathcal{S}}\subseteq P(X)$. In general, there’s no reason to believe that such a $\sigma$-algebra would be all of $P(X)$, and so there’s some interesting structure worth investigating. Our intuition is amazingly bad at conceiving of just how bizarre and perverse a “general” subset of the real line can be, and so the facts we learn about Lebesgue measurable sets seem sort of opaque when we first see them, but they will turn out to be useful.

First, let’s recall where they come from. There’s an outer measure $\mu^*$ associated with Lebesgue measure $\bar{\mu}$, and then there is the collection of sets measurable by $\mu^*$. These are the sets $E$ so that for any subset $A\subseteq\mathbb{R}$ we have

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

If $E$ is a Lebesgue measurable set of positive, finite measure, and if $\alpha\in\left[0,1\right)$ is a real number, then there exists some open interval $U$ so that

$\displaystyle\bar{\mu}(E\cap U)>\alpha\mu(U)$

That is, Lebesgue measurable sets do a pretty good job of covering sections of the real line. If we think of $\alpha$ as a measure of efficiency, we can always find some interval so that $E$ covers at least that portion of it.

Let $\mathcal{U}$ is the class of open sets. We already know that

$\displaystyle\bar{\mu}(E)=\inf\{\mu(U)\vert U\in\mathcal{U},E\subseteq U\}$

So we can find some open $U_0$ containing $E$ and satisfying $\alpha\mu(U_0)\leq\bar{\mu}(E)$. This will be the disjoint union of a sequence of open intervals $\{U_i\}$, and so it follows that

$\displaystyle\alpha\sum\limits_{i=1}^\infty\mu(U_i)\leq\sum\limits_{i=1}^\infty\bar{\mu}(E\cap U_i)$

We must have $\mu(U_i)\leq\bar{\mu}(E\cap U_i)$ for at least one of these summands, and this gives us the $U$ we were looking for.

If $E$ is again a Lebesgue measurable set with positive measure, then there exists an open interval containing ${0}$ and entirely contained in the set $D(E)$ of “differences” in $E$. That is, the set defined by

$\displaystyle D(E)=\{x-y\vert x,y\in E\}$

If $E$ contains an open interval $(a,b)$ itself, then this is trivially true. We can just take the desired interval to be $(a-b,b-a)$, which clearly falls into the difference set $D(E)$. But general Lebesgue measurable sets don’t contain open intervals. And yet this is not just a collection of scattered points, because then it would have measure zero. We have points clustered closely enough to fill up area, but not enough to actually fill out any open intervals. This is, in fact, supremely weird if you really think about it.

But even if $E$ doesn’t fill up an interval, it comes close! The above result tells us that we can find a $U$ so that

$\displaystyle\bar{\mu}(E\cap U)\geq\frac{3}{4}\mu(U)$

Now pick some $x\in\left(-\frac{1}{2}\mu(U),\frac{1}{2}\mu(U)\right)$. Take the set $E\cap U$, slide it over by adding $x$ to every point in the set, and call the result $(E\cap U)+x$. The set $(E\cap U)\cup((E\cap U)+x)$ is now contained in an open interval — $U\cup(U+x)$ — whose length is less than $\frac{3}{2}\mu(U)$. If $E\cap U$ and $(E\cap U)+x$ were disjoint, then they’d have to have the same measure, and their union would have measure greater than $\frac{3}{2}\mu(U)$ and wouldn’t fit into this interval. So $E\cap U$ and $(E\cap U)+x$ must overlap a bit.

But this tells us that there’s a point $a\in(E\cap U)$ that’s equal to the point $b+x\in((E\cap U)+x)$. Thus $a-b=x$ is a point in $D(E)$. And so $D(E)$ contains the whole interval $\left(-\frac{1}{2}\mu(U),\frac{1}{2}\mu(U)\right)$, as we wanted.

April 23, 2010 Posted by | Analysis, Measure Theory | 1 Comment

Lebesgue Measure and Affine Transformations

This didn’t make it up yesterday, due to some preoccupation on my part. Playing catch-up…

The thing that makes Lebesgue measure really special is the way that every part of the real line “looks like” every other part, and the same is true as we scale the line up or down. Once we specify the measure of one interval then the measure of every other interval — and every Lebesgue measurable set — is completely determined. But first, let’s show that this nice behavior actually holds.

Let $T$ be an affine transformation of the real line, defined by $T(x)=\alpha x+\beta$ for some real $\alpha\neq0$ and any real $\beta$. This is invertible, and the inverse function is affine as well: $T^{-1}(x)=\frac{x}{\alpha}-\frac{\beta}{\alpha}$. Given a subset $E\subseteq\mathbb{R}$, we write $T(E)$ for the image of $E$ under the transformation $T$$T(E)=\{\alpha x+\beta\vert x\in E\}$. I say that the outer and inner Lebesgue measures are both nicely behaved under the transformation $T$.

$\displaystyle\mu^*(T(E))=\lvert\alpha\rvert\mu^*(E)$
$\displaystyle\mu_*(T(E))=\lvert\alpha\rvert\mu_*(E)$

We’ll start by assuming that $\alpha>0$. In the other case, $T$ is the composition of the mapping $x\mapsto-x$ and an affine transformation with a positive multiplier, so we’ll have to come back and show that this reflection preserves our measures.

Let $T(\mathcal{S})$ be the collection of all sets of the form $T(E)$ with $E\in\mathcal{S}$. It should be clear that this is a $\sigma$-ring, and we will show that $T(\mathcal{S})=\mathcal{S}$. Indeed, if $E\in\mathcal{P}$ is a semiclosed interval $\left[a,b\right)$, then $E=T(F)$, where $F$ is the semiclosed interval $\left[\frac{a-\beta}{\alpha},\frac{b-\beta}{\alpha}\right)$. And so $E\in T(\mathcal{S})$, and the whole $\sigma$-ring $\mathcal{S}$ generated by $\mathcal{P}$ must be contained in $T(\mathcal{S})$: $\mathcal{S}\subseteq T(\mathcal{S})$. We can apply the exact same reasoning to $T^{-1}$ to conclude that $\mathcal{S}\subseteq T^{-1}(\mathcal{S})$, and thus $T(\mathcal{S})\subseteq\mathcal{S}$. This shows that $\mathcal{S}=T(\mathcal{S})$.

For every Borel set $E$, we define $\mu_1(E)=\mu(T(E))$ and $\mu_2(E)=\alpha\mu(E)$. These are both measures on $\mathcal{S}$. If $\mathcal{E}=\left[a,b\right)$ is a semiclosed interval, we calculate

\displaystyle\begin{aligned}\mu_1(E)&=\mu(T(E))\\&=\mu\left(\left[\alpha a+\beta,\alpha b+\beta\right)\right)\\&=(\alpha b+\beta)-(\alpha a\beta)\\&=\alpha(b-a)\\&=\alpha\mu(E)\\&=\mu_2(E)\end{aligned}

so the two measures agree on $\mathcal{P}$. But our extension theorem tells us that a measure on $\mathcal{S}$ is uniquely determined by its values on $\mathcal{P}$. And thus our assertion holds for Lebesgue measure on Borel sets.

Now if we apply these considerations to $T^{-1}$, we can show that

\displaystyle\begin{aligned}\mu^*(T(E))&=\inf\left\{\mu(F)\vert F\in\mathcal{S},E\subseteq F\right\}\\&=\inf\left\{\alpha\mu(T^{-1}(F))\vert T^{-1}(F)\in\mathcal{S},E\subseteq T^{-1}(F)\right\}\\&=\alpha\inf\left\{\mu(G)\vert G\in\mathcal{S},E\subseteq G\right\}\\&=\alpha\mu^*(E)\end{aligned}

Replacing $\inf$ by $\sup$ and $\subseteq$ by $\supseteq$, we also reach our conclusion for inner Lebesgue measure. But we still need to verify that $T$ preserves Lebesgue measurability in the first place. If $E$ is Lebesgue measurable, and $A$ is any set, then

\displaystyle\begin{aligned}\mu^*(A\cap T(E))+\mu^*(A\cap T(E)^c)&=\mu^*(T(T^{-1}(A)\cap E))+\mu^*(T(T^{-1}(A)\cap E^c))\\&=\alpha\left(\mu^*(T^{-1}(A)\cap E)+\mu^*(T^{-1}(A)\cap E^c)\right)\\&=\alpha\mu^(T^{-1}(A))\\&=\mu^*(A)\end{aligned}

and thus $T(E)$ is also Lebesgue measurable.

Now let’s go back and consider what happens with the reflection $x\mapsto-x$. This sends the semiclosed interval $\left[a,b\right)$ to the interval $\left(-b,-a\right]$. But this is a Borel set: $\left(-b,-a\right]=\left[-b,-a\right)\cup\{-a\}\setminus\{-b\}$, and the singletons are Borel sets. Thus we see that the reflection sends Borel sets to Borel sets. It should also be clear that it preserves Lebesgue measure, for $(-a)-(-b)=b-a$. If $D$ is a subset of a Borel set of measure zero, then the reflection of $D$ is again such a subset, and so we see that reflection preserves the Lebesgue measurable sets as well as their inner and outer measures.

April 22, 2010

Borel Sets and Lebesgue Measure

Let’s consider some of the easy properties of the Borel sets and Lebesgue measure we introduced yesterday.

First off, every countable set of real numbers is a Borel set of measure zero. In particular, every single point $\{a\}$ is a Borel set. Indeed, $\{a\}$ can be written as the countable intersection

$\displaystyle\{a\}=\bigcap\limits_{n=1}^\infty\left[a,a+\frac{1}{n}\right)$

so it’s a Borel set. Further, monotonicity tells us that

$\displaystyle\mu\left(\{a\}\right)=\lim\limits_{n\to\infty}\mu\left(\left[a,a+\frac{1}{n}\right)\right)=\lim\limits_{n\to\infty}\frac{1}{n}=0$

and so the singleton $\{a\}$ has measure zero. But $\mu$ is countably additive, so given any countable collection $A\subseteq\mathbb{R}$ the measure $\mu(A)$ is the sum of the measures of the individual points, each of which is zero.

Next, as I said when I introduced semiclosed intervals, we could have started with open intervals, but the details would have been messier. Now we can see that the $\sigma$-ring $\mathcal{S}$ generated by the collection $\mathcal{P}$ of semiclosed intervals is the same as that generated by the collection $\mathcal{U}$ of all open sets.

We can see, in particular, that each open interval $\left(a,b\right)$ is a Borel set. Indeed, the point $\{a\}$ is a Borel set, as is the semiclosed interval $\left[a,b\right)$, and we have the relation $\left(a,b\right)=\left[a,b\right)\setminus\{a\}$. Every other open set in $\mathbb{R}$ is a countable union of open intervals, and so they’re all Borel sets as well. Conversely, we could write

$\displaystyle\{a\}=\bigcap\limits_{n=1}^\infty\left(a-\frac{1}{n},a+\frac{1}{n}\right)$

and find the singleton $\{a\}$ in the $\sigma$-ring generated by $\mathcal{U}$. Then we can write $\left[a,b\right)=\left(a,b\right)\cup\{a\}$ and find every semiclosed interval in this $\sigma$-ring as well. And thus $\mathcal{S}=\mathcal{S}(\mathcal{P})\subseteq\mathcal{S}(\mathcal{U})$

We can also tie our current measure back to the concept of outer Lebesgue measure we introduced before. Back then, we defined the “volume” of a collection of open intervals to be the sum of the “volumes” of the intervals themselves. We defined the outer measure of a set to be the infimum of the volumes of finite open covers. And, indeed, this is exactly the outer measure $\mu^*$ corresponding to Lebesgue measure $\mu$.

Remember that the outer measure $\mu^*(E)$ is defined for a set $E\subseteq\mathbb{R}$ by

$\displaystyle\mu^*(E)=\inf\left\{\mu(F)\vert F\in\mathcal{S},E\subseteq F\right\}$

Since $\mathcal{U}\subseteq\mathcal{S}$, we have the inequality

$\displaystyle\mu^*(E)\leq\inf\left\{\mu(U)\vert U\in\mathcal{S},E\subseteq F\right\}$

On the other hand, if $\epsilon$ is any positive number, then by the definition of $\mu^*$ we can find a sequence $\left\{\left[a_n,b_n\right)\right\}$ of semiclosed intervals so that

$\displaystyle E\subseteq\bigcup\limits_{n=1}^\infty\left[a_n,b_n\right)$

and

$\displaystyle\sum\limits_{n=1}^\infty(b_n-a_n)\leq\mu^*(E)+\frac{\epsilon}{2}$

We can thus widen each of these semiclosed intervals just a bit to find

$\displaystyle E\subseteq\bigcup\limits_{n=1}^\infty\left(a_n-\frac{\epsilon}{2^{n+1}},b_n\right)=U\in\mathcal{U}$

and

$\displaystyle\mu(U)\leq\sum\limits_{i=1}^\infty(b_n-a_n)+\frac{\epsilon}{2}\leq\mu^*(E)+\epsilon$

Since $\epsilon$ was arbitrary, we find that $\mu(U)\leq\mu^*(E)$. And, thus, that

$\displaystyle\mu^*(E)=\inf\left\{\mu(U)\vert U\in\mathcal{S},E\subseteq F\right\}$

In effect, we’ve replaced the messily-defined “volume” of an open cover by the more precise Lebesgue measure $\mu$, but the result is the same. The “outer Lebesgue measure” from our investigations of multiple integrals is the same as the outer measure induced by our new Lebesgue measure.

April 20, 2010 Posted by | Analysis, Measure Theory | 5 Comments