The Unapologetic Mathematician

Mathematics for the interested outsider

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

   

Follow

Get every new post delivered to your Inbox.

Join 393 other followers