The Unapologetic Mathematician

Mathematics for the interested outsider

Extensions of Measures

Oops, forgot to post this this earlier…

We can put together what we’ve been doing recently to state the following theorem:

If \mu is a \sigma-finite measure on the ring \mathcal{R}, then there is a unique measure \bar{\mu} on the \sigma-ring \mathcal{S}(\mathcal{R}) extending \mu. That is, if E\in\mathcal{R}\subseteq\mathcal{S}(\mathcal{R}), then \bar{\mu}(E)=\mu(E). Further, the extended measure \bar{\mu} is also \sigma-finite.

The existence is straightforward. We can induce an outer measure, and then restrict it to get \bar{\mu}. It’s straightforward to verify from the definitions that \bar{\mu}(E)=\mu(E). And we know that since \mu is \sigma-finite, so is \mu^*, and thus \bar{\mu}.

What we need to show is that \bar{\mu} is unique. To this end, let \mu_1 and \mu_2 be two measures on \mathcal{S}(\mathcal{R}) that both extend \mu. Let \mathcal{M}\subseteq\mathcal{S}(\mathcal{R}) be the class of sets on which \mu_1 and \mu_2 agree; this obviously contains \mathcal{R}.

Now, if one of these two measures — say \mu_1 — is finite, and if \{E_i\}_{i=1}^\infty\subseteq\mathcal{M} is a monotone sequence of sets on which \mu_1 and \mu_2 agree, then the limit of this sequence is again in \mathcal{M}. Indeed, since measures are continuous, we must have


and similarly for \mu_2. Since \mu_1 is finite, and \mu_2 agrees with \mu_1 on \mathcal{M}, we have a sequence of finite measures \mu_1(E_i)=\mu_2(E_i). The limits of these sequences must then agree, and so \lim_iE_i\in\mathcal{M} as well. Thus \mathcal{M} is a monotone class. Since it contains \mathcal{R}, it must contain \mathcal{S}(\mathcal{R}), and thus \mu_1=\mu_2.

On the other hand, neither measure may be finite. In this case, let A\in\mathcal{R} be some fixed set of finite measure. Now \mathcal{R}\cap A — the collection of intersections of sets in \mathcal{R} with A — is again a ring, and \mathcal{S}(\mathcal{R})\cap A is the smallest \sigma-ring containing it. Restricting \mu_1 and \mu_2 to \mathcal{S}(\mathcal{R})\cap A gives finite measures, and we can use the argument above.

Now every set E\in\mathcal{S}(\mathcal{R}) can be covered by a countable, pairwise disjoint collection of sets A_i\in\mathcal{R}. For each one, we have E_i=E\cap A_i\in\mathcal{S}(\mathcal{R})\cap A_i, and so we must find \mu_1(E_i)=\mu_2(E_i). From here, countable additivity finishes the theorem.

In light of the uniqueness of this extension, we will just call the extended measure \mu again, rather than \bar{\mu}.

April 6, 2010 - Posted by | Analysis, Measure Theory


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