The Unapologetic Mathematician

Mathematics for the interested outsider

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


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