The Unapologetic Mathematician

Mathematics for the interested outsider

Signed Measures and Sequences

We have a couple results about signed measures and certain sequences of sets.

If \mu is a signed measure and \{E_n\} is a disjoint sequence of measurable sets so that the measure of their disjoint union is finite:

\displaystyle\left\lvert\mu\left(\biguplus\limits_{n=1}^\infty E_n\right)\right\rvert=\left\lvert\sum\limits_{n=1}^\infty\mu(E_n)\right\rvert<\infty

then the series

\displaystyle\sum\limits_{n=1}^\infty\mu(E_n)

is absolutely convergent. We already know it converges since the measure of the union is finite, but absolute convergence will give us all sorts of flexibility to reassociate and rearrange our series.

We want to separate out the positive and the negative terms in this series. We write E_n^+=E_n if \mu(E_n)\geq0 and E_n^+=\emptyset otherwise. Similarly, we write E_n^-=E_n if \mu(E_n)\leq0 and E_n^-=\emptyset otherwise. Then we write the two series

\displaystyle\begin{aligned}\mu\left(\biguplus\limits_{n=1}^\infty E_n^+\right)&=\sum\limits_{n=1}^\infty\mu(E_n^+)\\\mu\left(\biguplus\limits_{n=1}^\infty E_n^-\right)&=\sum\limits_{n=1}^\infty\mu(E_n^-)\end{aligned}

The terms of each series have a constant sign — positive for the first and negative for the second — and so if they diverge they can only diverge definitely — to \infty in the first case and to -\infty in the second. But at least one must converge or else we’d have \mu obtaining both infinite values. But the sum of all the E_n converges, and so both series must converge — if the series of \mu(E_n^+) diverged to \infty and the seris of \mu(E_n^-) converged, then there wouldn’t be enough negative terms in the series of \mu(E_n) for the whole thing to converge. But then since the positive terms and the negative terms both converge, the whole series is absolutely convergent.

Now we turn to some continuity properties. If \{E_n\} is a monotone sequence — if it’s decreasing we also ask that at least one \lvert\mu(E_n)\rvert<\infty — then

\displaystyle\mu\left(\lim\limits_{n\to\infty}E_n\right)=\lim\limits_{n\to\infty}\mu(E_n)

The proofs of both of these facts are exactly the same as for measures, except we need the monotonicity result from the end of yesterday’s post to be sure that once we hit one finite \mu(E_n), all the later \mu(E_m) will stay finite.

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June 23, 2010 - Posted by | Analysis, Measure Theory

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