The Unapologetic Mathematician

Mathematics for the interested outsider

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}.

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