# The Unapologetic Mathematician

## 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