The Unapologetic Mathematician

Mathematics for the interested outsider

Lebesgue Measure

So we’ve identified a measure on the ring \mathcal{R} of finite disjoint unions of semiclosed intervals. Now we want to apply our extension and completion theorems.

The smallest \sigma-ring \mathcal{S} containing \mathcal{R} is also the smallest one containing the collection \mathcal{P} of semiclosed intervals. As it turns out, it’s also a \sigma-algebra. Indeed, we can write the whole real line \mathbb{R} as the countable disjoint union of elements of \mathcal{P}.

\displaystyle\mathbb{R}=\bigcup\limits_{i=-\infty}^\infty\left[i,i+1\right)

and so \mathbb{R} itself must be in \mathcal{S}. We call \mathcal{S} the \sigma-algebra of “Borel sets” of the real line.

Our measure \mu — defined on elements of \mathcal{P} by \mu(\left[b,a\right))=b-a — is not just \sigma-finite, but actually finite on \mathcal{R}. And thus its extension to \mathcal{S} will still be \sigma-finite. The above decomposition of \mathbb{R} into a countable collection of sets of finite \mu-measure shows us that the extended measure is, in fact, totally \sigma-finite.

But our measure might not be complete. As the smallest \sigma-algebra containing \mathcal{P}, \mathcal{S} might not contain all subsets of sets of \mu-measure zero. And thus we form the completions \overline{\mathcal{S}} of our \sigma-algebra and \bar{\mu} of our measure. We call \overline{\mathcal{S}} the \sigma-algebra of “Lebesgue measurable sets”, and \bar{\mu} is “Lebesgue measure” (remember, it’s pronounced “luh-BAYG”). In fact, the incomplete measure \mu on Borel sets is also often called Lebesgue measure.

April 19, 2010 Posted by | Analysis, Measure Theory | 9 Comments

An Example of a Measure

At last we can show that the set function we defined on semiclosed intervals is a measure. It’s clearly real-valued and non-negative. We already showed that it’s monotonic, and this will come in handy as we show that it’s countably additive.

So, if \{E_i\} is a countable disjoint sequence of semiclosed intervals whose union is also a semiclosed interval E, then our first monotonicity property shows that for any finite n we have

\displaystyle\sum\limits_{i=1}^n\mu(E_i)\leq\mu(E)

and so in the limit we must still have

\displaystyle\sum\limits_{i=1}^\infty\mu(E_i)\leq\mu(E)

But the sequence \{E_i\} covers E, and so our other monotonicity property shows that

\displaystyle\mu(E)\leq\sum\limits_{i=1}^\infty\mu(E_i)

which gives us the equality we want.

But this still isn’t quite a measure. Why not? It’s only defined on the collection \mathcal{P} of semiclosed intervals, and not on the ring \mathcal{R} of finite disjoint unions. But we’re in luck: there is a unique finite measure \bar{\mu} on \mathcal{R} extending \mu on \mathcal{P}. That is, if E\in\mathcal{P}, then \bar{\mu}(E)=\mu(E).

Every set in \mathcal{R} is a finite disjoint union of semiclosed intervals, but not necessarily uniquely. Let’s say we have both

\displaystyle E=\bigcup\limits_{i=1}^mE_i
\displaystyle E=\bigcup\limits_{j=1}^nF_j

Then for each i we have

\displaystyle E_i=\bigcup\limits_{j=1}^n(E_i\cap F_j)

which represents E_i\in\mathcal{P} as a finite disjoint union of other sets in \mathcal{P}. Since \mu is finitely additive, we must have

\displaystyle\sum\limits_{i=1}^m\mu(E_i)=\sum\limits_{i=1}^m\sum\limits_{j=1}^n\mu(E_i\cap F_j)

and, similarly

\displaystyle\sum\limits_{j=1}^n\mu(F_j)=\sum\limits_{j=1}^n\sum\limits_{i=1}^m\mu(E_i\cap F_j)

But since these sums are finite we can switch their order with no trouble. Thus we can unambiguously define

\bar{\mu}(E)=\sum\limits_{i=1}^m\mu(E_i)

which doesn’t depend on how we represent E as a finite disjoin union of semiclosed intervals.

This function \bar{\mu} clearly extends \mu, since if E\in\mathcal{P} we can just use E itself as our finite disjoint union. It’s also easily seen to be finitely additive, and that there’s not really any other way to define a finitely additive set function to extend \mu. But we still need to show countable additivity.

So, let \{E_i\} be a disjoint sequence of sets in \mathcal{R} whose union E is also in \mathcal{R}. Then for each i we have

\displaystyle E_i=\bigcup\limits_{j=1}^{n_j}E_{ij}
\displaystyle\bar{\mu}(E_i)=\sum\limits_{j=1}^{n_j}\mu(E_{ij})

If E happens to be in \mathcal{P}, then the collection of all the E_{ij} is countable and disjoint, and we can use the countable additivity of \mu we proved above to show

\displaystyle\bar{\mu}(E)=\mu(E)=\sum\limits_{i=1}^\infty\sum\limits_{j=1}^{n_j}\mu(E_{ij})=\sum\limits_{i=1}^\infty\bar{\mu}(E_i)

In general, though, E is a finite disjoint union

\displaystyle E=\bigcup\limits_{k=1}^nF_k

and we can apply the previous result to each of the F_k:

\displaystyle\begin{aligned}\bar{\mu}(E)&=\sum\limits_{k=1}^n\bar{\mu}(F_k)\\&=\sum\limits_{k=1}^n\sum\limits_{i=1}^\infty\bar{\mu}(E_i\cap F_k)\\&=\sum\limits_{i=1}^\infty\sum\limits_{k=1}^n\bar{\mu}(E_i\cap F_k)\\&=\sum\limits_{i=1}^\infty\bar{\mu}(E_i)\end{aligned}

From here on out, we’ll just write \mu instead of \bar{\mu} for this measure on \mathcal{R}.

April 16, 2010 Posted by | Analysis, Measure Theory | 1 Comment

An Example of Monotonicity

We continue with our example and show that the set function \mu which assigns any semiclosed interval its length has various monotonicity properties.

First off, let \{E_1,\dots,E_n\} be a finite, disjoint collection of semiclosed intervals, all of which are contained in another semiclosed interval E. Then we have the inequality

\displaystyle\sum\limits_{i=1}^n\mu(E_i)\leq\mu(E)

Indeed, we can write E=\left[a,b\right), E_i=\left[a_i,b_i\right), and without loss of generality assume that a_1<a_2<\dots<a_n. Then our hypotheses tell us that

\displaystyle a\leq a_1<b_1\leq a_2<b_2\leq\dots\leq a_n<b_n\leq b

and thus

\displaystyle\begin{aligned}\sum\limits_{i=1}^n\mu(E_i)&=\sum\limits_{i=1}^n(b_i-a_i)\\&\leq\sum\limits_{i=1}^n(b_i-a_i)+\sum\limits_{i=1}^{n-1}(a_{i+1}-b_i)\\&=b_n-a_1\leq b-a=\mu(E)\end{aligned}

On the other hand, if F=\left[a,b\right] is a closed interval contained in the union of a finite number of bounded open intervals U_i=\left(a_i,b_i\right), then we have the strict inequality

\displaystyle b-a<\sum\limits_{i=1}^n(b_i-a_i)

We can rearrange the open intervals by picking U_1 to contain a. Then if b_1>b we have F\subseteq U_1 and we can discard all the other sets since they only increase the right hand side of the inequality. But if b_1\leq b, we can pick some U_2 containing b_1. Now we repeat, asking whether b_2 is greater or less than b. Eventually we’ll have a finite collection of U_i satisfying a_1<a<b_1, a_n<b<b_n, and a_{i+1}<b_i<b_{i+1}. It follows that

\displaystyle\begin{aligned}b-a&<b_n-a_1\\&=(b_1-a_1)+\sum\limits_{i=1}^{n-1}(b_{i+1}-b_i)\\&\leq\sum\limits_{i=1}^n(b_i-a_i)\end{aligned}

What does this have to do with semiclosed intervals? Well, if \{E_i\}_{i=1}^\infty is a countable sequence of semiclosed intervals that cover another semiclosed interval E, then we have the inequality

\displaystyle\mu(E)\leq\sum\limits_{i=1}^\infty\mu(E_i)

If E=\emptyset, then this is trivially true, so we’ll assume it isn’t, and let \epsilon be a positive number with \epsilon<b-a. Then we have the closed set F=\left[a,b-\epsilon\right]. We can also pick any positive number \delta and define U_i=\left(a_i-\frac{\delta}{2^i},b_i\right).

Now F is smaller than E, and each U_i is larger than the corresponding E_i, and so we find that F is a closed interval covered by the open intervals U_i. But the Heine-Borel theorem says that F is compact, and so we can find a finite collection of the U_i which cover F. Renumbering the open intervals, we have

\displaystyle F\subseteq\bigcup\limits_{i=1}^nU_i

and our above result tells us that

\displaystyle\begin{aligned}\mu(E)-\epsilon&=b-a-\epsilon\\&<\sum\limits_{i=1}^n\left(b_i-a_i+\frac{\delta}{2^i}\right)\\&\leq\sum\limits_{i=1}^\infty\mu(E_i)+\delta\end{aligned}

Since we can pick \epsilon and \delta to be arbitrarily small, the desired inequality follows.

April 15, 2010 Posted by | Analysis, Measure Theory | 1 Comment

Semiclosed Intervals

Before we go any further, let’s work towards an actual example of a measure. This one, in the long run, will be useful to us.

The underlying space X we’re interested in is the real line. We need to start with a class \mathcal{P} of sets we’re interested in measuring. Specifically, we’re going to take \mathcal{P} to be the class of finite intervals, open on the right and closed on the left. That is, given finite real numbers a<b we consider the interval

\displaystyle\left[a,b\right)=\{x\vert a\leq x<b\}

Such a bounded interval we’ll call “semiclosed”. We’ll also throw \emptyset into \mathcal{P} and let this count as a degenerate sort of semiclosed interval.

Now, given two semiclosed intervals, their intersection is again a semiclosed interval. One possibility is that one interval contains the other, in which case the intersection is the smaller interval. Another possibility is that the intervals are disjoint, in which case their intersection is empty. The last possibility is that they overlap: we consider \left[a,b\right) and \left[c,d\right) with a<c<b<d. Then their intersection is \left[c,b\right), which is a semiclosed interval.

The difference of two semiclosed intervals may or may not be a semiclosed interval. If intervals overlap, as above, then \left[a,b\right)\setminus\left[c,d\right)=\left[a,c\right), and \left[c,d\right)\setminus\left[a,b\right)=\left[b,d\right). If the intervals are disjoint, then the difference is just the original interval. But if \left[a,b\right) contains \left[c,d\right), then the difference is \left[a,c\right)\cup\left[d,b\right). This isn’t a semiclosed interval, but it’s a finite disjoint union of semiclosed intervals.

But we know that these properties are exactly what we need to show that the collection \mathcal{R} of finite disjoint unions of intervals in \mathcal{P} is a ring. We could have started with open intervals or closed intervals, but then we wouldn’t have such a nice ring pop out.

We will define a finite set function \mu:\mathcal{P}\rightarrow\mathcal{R}. For an interval \left[a,b\right), we define \mu(\left[a,b\right))=b-a. For the empty set, we define \mu(\emptyset)=0. This is the function that will be developed into our measure.

April 14, 2010 Posted by | Analysis, Measure Theory | 7 Comments

Using Measurable Covers and Kernels II

Following yesterday’s post, here are some more useful facts that we can prove with the help of measurable covers and measurable kernels.

If E and F are disjoint sets in \mathcal{H}(\mathcal{S}), then

\displaystyle\mu_*(E\cup F)\leq\mu_*(E)+\mu^*(F)\leq\mu^*(E\cup F)

Here, we take A to be a measurable cover of F and B to be a measurable kernel of E\cup F. The difference B\setminus A must be contained in E, and so

\displaystyle\mu_*(E\cup F)=\mu(B)=\mu(B\setminus A)+\mu(A)\leq\mu_*(E)+\mu^*(F)

On the other hand, we can take A to be a measurable kernel of E and B to be a measurable cover of E\cup F. Now the difference B\setminus A is contained in F, and we find

\displaystyle\mu^*(E\cup F)=\mu(B)=\mu(A)+\mu(B\setminus A)\geq\mu_*(E)+\mu^*(F)

Now, if E\in\overline{\mathcal{S}}, then for every A\subseteq X whatsoever, we have

\displaystyle\mu_*(A\cap E)+\mu^*(A^c\cap E)=\bar{\mu}(E)

We can take A\cap E and A^c\cap E and stick them into the previous result to find

\displaystyle\begin{aligned}\mu_*(E)&=\mu_*((A\cap E)\cup(A^c\cap E))\\&\leq\mu_*(A\cap E)+\mu^*(A^c\cap E)\\&\leq\mu^*((A\cap E)\cup(A^c\cap E))\\&=\mu^*(E)\end{aligned}

But since E\in\overline{\mathcal{S}}, we know that \mu_*(E)=\mu^*(E)=\bar{\mu}(E), and this establishes our result.

Interestingly, we can use this method of inner measures as an alternative approach to our extension theorems. If \mu is a \sigma-finite measure on a ring \mathcal{R}, and if \mu^* is the induced outer measure on \mathcal{H}(\mathcal{R}), then for every set E\in\mathcal{R} of finite measure and every A\in\mathcal{H}(\mathcal{R}) we have

\displaystyle\mu_*(A\cap E)=\mu(E)-\mu^*(A^c\cap E)

Then if E and F are two sets in \mathcal{R} such that A\cap E=A\cap F, then we find

\mu(E)-\mu^*(A^c\cap E)=\mu(F)-\mu^*(A^c\cap F)

and so we can use this formula as the definition of the inner measure \mu_*. Then we can define a set E\in\mathcal{H}(\mathcal{R}) with \mu^*(E)<\infty to be \mu^*-measurable if the inner and outer measures match: \mu_*(E)=\mu^*(E). And from here, the rest of the theory is as before.

April 13, 2010 Posted by | Analysis, Measure Theory | 2 Comments

Using Measurable Covers and Kernels I

There are a bunch of useful facts that we can prove with the help of measurable covers and measurable kernels. These will allow us to move statements we want to show about outer and inner measures to the realm of proper measures, where we can use nice things like additivity.

For example, given a pairwise disjoint sequence \{E_i\}\subseteq\mathcal{H}(\mathcal{S}), we can show that

\displaystyle\mu_*\left(\bigcup\limits_{i=1}^\infty E_i\right)\geq\sum\limits_{i=1}^\infty\mu_*(E_i)

Just pick a measurable kernel F_i for each E_i. Then we can use countable additivity to show

\displaystyle\sum\limits_{i=1}^\infty\mu_*(E_i)=\sum\limits_{i=1}^\infty\mu(F_i)=\mu\left(\bigcup\limits_{i=1}^\infty F_i\right)\leq\mu_*\left(\bigcup\limits_{i=1}^\infty E_i\right)

For another, given a set A\in\mathcal{H}(\mathcal{S}) and a disjoint sequence \{E_i\}\subseteq\overline{\mathcal{S}} whose union is E, then

\displaystyle\mu_*(A\cap E)=\sum\limits_{i=1}^\infty\mu_*(A\cap E_i)

This time, let F be a measurable kernel of A\cap E, so that

\displaystyle\mu_*(A\cap E)=\mu(F)=\sum\limits_{i=1}^\infty\bar{\mu}(F\cap E_i)\leq\sum\limits_{i=1}^\infty\mu_*(A\cap E_i)

On the other hand, A\cap E is the union of the A\cap E_i, and so we can use the previous result to get the opposite inequality.

Next: if E\in\overline{\mathcal{S}}, then clearly \mu_*(E)=\mu^*(E)=\bar{\mu}(E). But conversely, if \mu_*(E)=\mu^*(E)<\infty, then E\in\overline{\mathcal{S}}. To see this, let A be a measurable kernel and B be a measurable cover of E. Then we calculate

\displaystyle\mu(B\setminus A)=\mu(B)-\mu(A)=\mu^*(E)-\mu_*(E)=0

But E\setminus A\subseteq B\setminus A, so E\setminus A\in\overline{\mathcal{S}} (by the completeness of \bar{\mu}), and thus E=(E\setminus A)\cup A\in\overline{\mathcal{S}}. Thus sets of finite measure in \overline{\mathcal{S}} are exactly those in \mathcal{H}(\mathcal{S}) for which the outer and inner measures coincide.

Interestingly, notice how the last step of this proof echoes our earlier result that a set is Jordan measurable if and only if the Jordan content of its boundary is zero.

April 12, 2010 Posted by | Analysis, Measure Theory | 1 Comment

Measurable Kernels

A measurable kernel is the flip side of a measurable cover. Specifically, given E\in\mathcal{H}(\mathcal{S}), a measurable kernel of E is a set F\in\mathcal{S} such that F\subseteq E, and if for every G\in\mathcal{S} with G\subseteq E\setminus F we have \mu(G)=0. And, as it happens, every set E\in\mathcal{H}(\mathcal{S}) has a measurable kernel.

To find it, let \hat{E} be a measurable cover of E. Then let N be a measurable cover of \hat{E}\setminus E, and set F=\hat{E}\setminus N\in\mathcal{S}. Since N contains \hat{E}\setminus E, we find

\displaystyle F=\hat{E}\setminus N\subseteq\hat{E}\setminus(\hat{E}\setminus E)=E.

If G\subseteq E\setminus F, then

\displaystyle G\subseteq E\setminus(\hat{E}\setminus N)=E\cap N\subseteq N\setminus(\hat{E}\setminus E)

Since N was picked to be a measurable cover of \hat{E}\setminus E, we conclude that \mu(G)=0, as we hoped.

Now if F is a measurable kernel of E, then \mu(F)=\mu_*(E). Indeed, since F\subseteq E, we have \mu(F)\leq\mu_*(E). If this inequality is strict then \mu(F)<\infty, and there must be some F_0\in\mathcal{S} with F_0\subseteq E and \mu(F_0)>\mu(F). But F_0\setminus F\subseteq E\setminus F, while \mu(F_0\setminus F)\geq\mu(F_0)-\mu(F)>0, contradicting the fact that F was chosen to be a measurable kernel of E.

The symmetric difference of any two measurable kernels is negligible. Given two measurable kernels F_1 and F_2, we know that F_1\subseteq F_1\cup F_2\subseteq E. This implies that (F_1\cup F_2)\setminus F_1\subseteq E\setminus F_1, and thus \mu((F_1\cup F_2)\setminus F_1)=0. Similarly, \mu((F_1\cup F_2)\setminus F_2)=0, and thus \mu(F_1\Delta F_2)=0.

April 10, 2010 Posted by | Analysis, Measure Theory | 2 Comments

Inner Measures

A quick one for today.

In analogy with the outer measure \mu^* induced on the hereditary \sigma-ring \mathcal{H}(\mathcal{S}) by the measure \mu on the \sigma-ring \mathcal{S}, we now define the “inner measure” \mu_* that \mu induces on the same hereditary \sigma-ring \mathcal{H}(\mathcal{S}). We’ve seen that the outer measure is

\displaystyle\mu^*(E)=\inf\{\mu(F)\vert E\subseteq F\mathcal{S}\}

Accordingly, the inner measure is a set function defined by

\displaystyle\mu_*(E)=\sup\{\mu(F)\vert E\supseteq F\mathcal{S}\}

In a way, the properties of \mu_* are “dual” to those of \mu^*. The easy ones are the same: it’s non-negative, monotone, and \mu_*(0)=0.

We could also define \mu_* in terms of the completed measure. Since \mathcal{S}\subseteq\overline{\mathcal{S}}, it’s clear

\displaystyle\mu_*(E)=\sup\{\mu(F)\vert E\supseteq F\mathcal{S}\}\leq\sup\{\bar{\mu}(F)\vert E\supseteq F\overline{\mathcal{S}}\}

On the other hand, the definition of the completion says that for every F\in\overline{\mathcal{S}} there is a G\in\mathcal{S} with G\subseteq F and \mu(G)=\bar{\mu}(F), and so this is actually an equality.

April 8, 2010 Posted by | Analysis, Measure Theory | 3 Comments

Approximating Sets of Finite Measure

So, we’ve got a \sigma-finite measure \mu on a ring \mathcal{R}, and we extend it to a measure on the \sigma-ring \mathcal{S}(\mathcal{R}). But often it’s a lot more convenient to work with \mathcal{R} itself than the whole of \mathcal{S}(\mathcal{R}. So, to what extent can we do this efficiently?

As it turns out, if E\in\mathcal{S}(\mathcal{R} has finite measure and \epsilon>0, then we can find a set E_0\in\mathcal{R} so that \mu(E\Delta E_0)\leq\epsilon.

Any set E\in\mathcal{S}(\mathcal{R}) can be covered by a sequence of sets in \mathcal{R}, and we know that

\displaystyle\inf\left\{\sum\limits_{i=1}^\infty\mu(E_i)\bigg\vert E\subseteq\bigcup\limits_{i=1}^\infty E_i, \{E_i\}\subseteq\mathcal{R}\right\}

That is, we can find such a cover satisfying

\displaystyle\mu\left(\bigcup\limits_{i=1}^\infty E_i\right)\leq\mu(E)+\frac{\epsilon}{2}

But since \mu is continuous, we see that

\displaystyle\mu\left(\bigcup\limits_{i=1}^\infty E_i\right)=\mu\left(\lim\limits_{n\to\infty}\bigcup\limits_{i=1}^nE_i\right)=\lim\limits_{n\to\infty}\mu\left(\bigcup\limits_{i=1}^nE_i\right)

The sequence of numbers increases until it’s within \frac{\epsilon}{2} of its limit. That is, there is some n so that if we define E_0 to be the union of the first n sets in the sequence, we have

\displaystyle\mu\left(\bigcup\limits_{i=1}^\infty E_i\right)\leq\mu(E_0)+\frac{\epsilon}{2}

But now we can find

\displaystyle\begin{aligned}\mu(E\setminus E_0)\leq\mu\left(\left(\bigcup\limits_{i=1}^\infty E_i\right)\setminus E_0\right)&=\mu\left(\bigcup\limits_{i=1}^\infty E_i\right)-\mu(E_0)\leq\frac{\epsilon}{2}\\\mu(E_0\setminus E)\leq\mu\left(\left(\bigcup\limits_{i=1}^\infty E_i\right)\setminus E\right)&=\mu\left(\bigcup\limits_{i=1}^\infty E_i\right)-\mu(E)\leq\frac{\epsilon}{2}\end{aligned}

And thus \mu(E\Delta E_0)\leq\epsilon.

April 7, 2010 Posted by | Analysis, Measure Theory | 1 Comment

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

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