# The Unapologetic Mathematician

## 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