The Unapologetic Mathematician

Mathematics for the interested outsider

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<n}E_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<n}E_i\right)\cap X_0\right)\Delta\left(E_n\cap X_0\right)\\&=\left(\left(E_n\cap X_0\right)\setminus\bigcup\limits_{1\leq i<n}\left(E_i\cap X_0\right)\right)\Delta F_n\\&=\left(F_n\setminus\bigcup\limits_{1\leq i<n}F_n\right)\Delta F_n\\&=F_n\Delta F_n=0\end{aligned}

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



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