The Unapologetic Mathematician

Mathematics for the interested outsider

Completions of Measures

We’ve shown that we can uniquely extend a \sigma-finite measure \mu on a ring \mathcal{R} to a unique \sigma-finite measure on the \sigma-ring \mathcal{S}(\mathcal{R}). But, of course, we actually found that we could restrict the outer measure \mu^* to the \sigma-ring \overline{\mathcal{S}} of \mu^*-measurable sets, which may be larger than \mathcal{S}(\mathcal{R}). Luckily, we can get this extra ground without having to go through the outer measure.

What’s the essential difference? What do we know about \overline{\mathcal{S}} that we don’t know about \mathcal{S}(\mathcal{R})? The measure on \overline{\mathcal{S}} is complete. The smaller \sigma-ring \mathcal{S}(\mathcal{R}) may not contain all negligible sets.

So, let’s throw them in; if \mu is a measure on a \sigma-ring \mathcal{S}, define \overline{\mathcal{S}} to be the class of all sets E\Delta N, where E\in\mathcal{S}, and N is a negligible with respect to the measure \mu, or “\mu-negligible”. This collection \overline{\mathcal{S}} is a \sigma-algebra, and the set function \bar{\mu} defined on \overline{\mathcal{S}} by \bar{\mu}(E\Delta N)=\mu(E) is a complete measure called the “completion” of \mu.

First, given sets E\in\mathcal{S} and N\subseteq A\in\mathcal{S} with \mu(A)=0, we have the two equations

\displaystyle\begin{aligned}E\cup N&=(E\setminus A)\Delta\left(A\cap(E\cup N)\right)\\E\Delta N&=(E\setminus A)\cup\left(A\cap(E\Delta N)\right)\end{aligned}

These tell us that any set that can be written as the symmetric difference of a set in \mathcal{S} and a measurable set can also be written as the union of two other such sets, and vice versa. That is, we can also characterize \overline{\mathcal{S}} as the class of sets of the form E\cup N instead of E\Delta N.

This characterization makes it clear that \overline{\mathcal{S}} is closed under countable unions. Indeed, just write each set in a sequence as a union of a set E_i from \mathcal{S} and a \mu-negligible set N_i. The countable union of the E_i is still in \mathcal{S}, and the countable union of the negligible sets is still negligible. Thus \overline{\mathcal{S}} is a \sigma-ring.

Now, if we have two ways of writing a set in \overline{\mathcal{S}}, say E_1\Delta N_1=E_2\Delta N_2, then we also have the equation E_1\Delta E_2=N_1\Delta N_2. And, therefore, \mu(E_1\Delta E_2)=\mu(N_1\Delta N_2)=0. Therefore, \mu(E_1)=\mu(E_2), and the above definition of \bar{\mu} is unambiguous.

Using the characterization of \overline{\mathcal{S}} in terms of unions, it’s easy to verify that \bar{\mu} is a measure. We only need to check countable additivity, which is perfectly straightforward; the union of a pairwise disjoint sequence of sets \{E_i\cup N_i\} is formed by taking the union of the E_i and the N_i. The measure \mu is countably additive on the E_i, and the union of the N_i is still negligible.

Finally, we must show that \bar{\mu} is complete. But if we have a set E\cup N with \bar{\mu}(E\Delta N)=0, then \mu(E)=0 and N\subseteq A with \mu(A)=0. Thus any subset M\subseteq E\cup N is also a subset of E\cup A, with \mu(E\cup A)=0. Writing it as \emptyset\Delta M, we find M\in\overline{\mathcal{S}}, and so \bar{\mu} is complete.

There’s just one fly in the ointment: we don’t really know that this complete measure is the same as the one we get by restricting from \mu^* to \mu^*-measurable sets. It turns out that if \mu is \sigma-finite on \mathcal{R}, then the completion of the extension of \mu to \mathcal{S}(\mathcal{R}) is the same as this restriction.

Since we’ve been using \overline{\mathcal{S}} for the completion, let’s write \mathcal{S}^* for the class of \mu^*-measurable sets. Since \mu^* restricted to \mathcal{S}^* is complete, it follows that \overline{\mathcal{S}}\subseteq\mathcal{S}^*, and that \bar{\mu} and \mu^* coincide on \overline{\mathcal{S}}. We just need to show that \mathcal{S}^*\subseteq\overline{\mathcal{S}}.

In light of the fact that \mu^* is also \sigma-finite on \mathcal{S}^*, we just need to show that if E\in\mathcal{S}^* has finite outer measure, then E\in\overline{\mathcal{S}}. But in this case E has a measurable cover F with \mu^*(F)=\mu(F)=\mu^*(E). Since these are finite, we find that \mu^*(F\setminus E)=0. But F\setminus E also has a measurable cover G, with \mu(G)=\mu^*(F\setminus E)=0. And so we can write E=(F\setminus G)\cup(E\cap G), showing that E\in\overline{\mathcal{S}}.

April 6, 2010 Posted by | Analysis, Measure Theory | 3 Comments

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

\displaystyle\mu_1\left(\lim\limits_{i\to\infty}E_i\right)=\lim\limits_{i\to\infty}\mu_1(E_i)

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

   

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