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


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