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 | 5 Comments

Regular Outer Measures

As usual, let \mathcal{R} be a ring of sets, \mathcal{S}(\mathcal{R}) be the smallest \sigma-algebra containing \mathcal{R}, and \mathcal{H}(\mathcal{R}) be the smallest hereditary \sigma-algebra containing \mathcal{R}. We’ve asked about the relation between a measure \mu on \mathcal{R}, the outer measure \mu^* it induces on \mathcal{H}(\mathcal{R}), and the measure \bar{\mu} we get by restricting \mu^* to \mathcal{S}(\mathcal{R}). But for now, let’s consider what happens when we start with an outer measure on \mathcal{H}(\mathcal{R}).

Okay, so we’ve got an outer measure \mu^* on a hereditary \sigma-ring \mathcal{H} — like \mathcal{H}(\mathcal{R}). We can define the \sigma-ring \overline{\mathcal{S}} of \mu^*-measurable sets and restrict \mu^* to a measure \bar{\mu} on \overline{\mathcal{S}}. And then we can turn around and induce an outer measure \bar{\mu}^* on the hereditary \sigma-ring \mathcal{H}(\overline{\mathcal{S}}).

Now, in general there’s no reason that these two should be related. But we have seen that if \mu^* came from a measure \mu (as described at the top of this post), then \mathcal{H}(\overline{\mathcal{S}})=\mathcal{H}(\mathcal{R}), and the measure \bar{\mu}^* induced by \bar{\mu} is just \mu^* back again!

When this happens, we say that \mu^* is a “regular” outer measure. And so we’ve seen that any outer measure induced from a measure on a ring is regular. The converse is true as well: if we have a regular outer measure \mu^*=\bar{\mu}^*, then it is induced from the measure \bar{\mu} on \overline{\mathcal{S}}. Induced and regular outer measures are the same.

Doesn’t this start to look a bit like a Galois connection?

April 2, 2010 Posted by | Analysis, Measure Theory | Leave a comment

Measurable Covers

Again, let \mathcal{R} be a ring, let \mathcal{S}(\mathcal{R}) be the smallest \sigma-ring containing \mathcal{R}, and let \mathcal{H}(\mathcal{R}) be the smallest hereditary \sigma-ring containing \mathcal{R}. Given a measure \mu on \mathcal{R}, it induces an outer measure \mu^* on \mathcal{H}(\mathcal{R}), which restricts to a measure \bar{\mu} on \mathcal{S}(\mathcal{R}).

The sets in E\in\mathcal{H}(\mathcal{R}) are easily described — they’re everything that can be countably covered by sets in \mathcal{R} — but we don’t have a measure on this collection. Instead, we need good approximations of these sets by sets in \mathcal{S}(\mathcal{R}), where we do have a measure. And so we say that a set F\in\mathcal{S}(\mathcal{R}) is a “measurable cover” of E if E\subseteq F, and if for every G\in\mathcal{S}(\mathcal{R}) with G\subseteq F\setminus E we have \bar{\mu}(G)=0. That is, F may be larger than E, but only negligibly.

If E\in\mathcal{H}(\mathcal{R}) has \sigma-finite outer measure, then it has a measurable cover F with \bar{\mu}(F)=\mu^*(E). Indeed, our hypothesis is that E can be written as the countable disjoint union of sets of finite outer measure, so if we can show this for sets of finite measure the \sigma-finite case will follow.

By the comparisons we showed last time, for every n there is some set F_n\in\mathcal{S}(\mathcal{R}) so that E\subseteq F_n and \bar{\mu}(F_n)\leq\mu^*(E)+\frac{1}{n}. Taking F to be the intersection of this sequence, we must have

\displaystyle \mu^*(E)\leq\bar{\mu}(F)\leq\bar{\mu}(F_n)\leq\mu^*(E)+\frac{1}{n}

Since n is arbitrary, we must have \bar{\mu}(F)=\mu^*(E). If G\in\mathcal{S}(\mathcal{R}) fits into F\setminus E, then E\subseteq F\setminus G. And so:

\bar{\mu}(F)=\mu^*(E)\leq\bar{\mu}(F\setminus G)=\bar{\mu}(F)-\bar{\mu}(G)\leq\bar{\mu}(F)

Since \bar{\mu}(F)<\infty we can subtract it off and find \bar{\mu}(G)=0.

In fact, any measurable cover F must have \bar{\mu}(F)=\mu^*(E). Further, any two of them have a negligible difference.

Indeed, if we have two measurable covers, F_1 and F_2, then E\subseteq F_1\cap F_2\supseteq F_1, which tells us that F_1\setminus(F_1\cap F_2)\subseteq F_1\setminus E. But since F_1 is a measurable cover, we conclude that \bar{\mu}(F_1\setminus(F_1\cap F_2))=0, and similarly that \bar{\mu}(F_2\setminus(F_1\cap F_2))=0. But F_1\Delta F_2 is the union of these two differences, and so \bar{\mu}(F_1\Delta F_2)=0.

If \mu^*(E)=\infty, then any measurable cover must have \bar{\mu}(F)=\infty as well. On the other hand, if \mu^*(E) is finite, then there exists at least one measurable cover with \bar{\mu}(F)=\mu^*(E). Then any other measurable cover differs negligibly, and so has the same measure.

And so if the measure \mu on \mathcal{R} is \sigma-finite, then so are the measures \bar{\mu} on \mathcal{S}(\mathcal{R}) and \bar{\mu} on \overline{\mathcal{S}}. We’ve already seen that \mu^* must be \sigma-finite in this situation, and so any \mu^*-measurable set can be covered by a countable sequence of \mu^*-finite sets. Applying the above results to each of the sets in this sequence gives our result.

April 1, 2010 Posted by | Analysis, Measure Theory | 4 Comments