# The Unapologetic Mathematician

## 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. Then our hypotheses tell us that $\displaystyle a\leq a_1

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_n, and $a_{i+1}. It follows that \displaystyle\begin{aligned}b-a&

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. 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