# The Unapologetic Mathematician

## 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