# The Unapologetic Mathematician

## Signed Measures

We continue what we started yesterday by extending the notion of a measure. We want something that captures the indefinite integrals of every function for which it’s defined.

And so we introduce a “signed measure”. This is essentially just like a measure, except we allow negative values as well. That is, $\mu$ is an extended real-valued, countably additive set function. But we want to prune the concept slightly.

First off, we insist that $\mu(\emptyset)=0$. Additivity tells us that $\mu(E)=\mu(E\uplus\emptyset)=\mu(E)+\mu(\emptyset)$; if there is any measurable set $E$ at all with $\mu(E)$ finite, then $\mu(\emptyset)=0$ follows, so our condition just rules out the degenerate cases where $\mu(E)=\infty$ or $\mu(E)=-\infty$ for all measurable $E$.

Secondly, we insist that $\mu$ can take only one of the values $\pm\infty$. That is, we can’t have one measurable $E$ with $\mu(E)=\infty$ and another measurable $F$ with $\mu(F)=-\infty$. Indeed, if this were the case then we’d have to deal with some indeterminate sums. We can’t quite be sure of this just considering $E\cup F$ since the two might not be disjoint, but we can consider the following three equations that follow from additivity:

\displaystyle\begin{aligned}\mu(E)&=\mu(E\setminus F)+\mu(E\cap F)\\\mu(F)&=\mu(F\setminus E)+\mu(E\cap F)\\\mu(E\Delta F)&=\mu(E\setminus F)+\mu(F\setminus E)\end{aligned}

Either $\mu(E\cap F)$ is finite or it’s not. If it’s finite, then the first two equations tell us that $\mu(E\setminus F)=\infty$ and $\mu(F\setminus E)=-\infty$, and so the sum in the third equation is indeterminate. On the other hand, if $\mu(E\cap F)=\infty$ then the sum in the second equation must be indeterminate to satisfy $\mu(F)=-\infty$, and similarly the sum in the first equation would have to be indeterminate if $\mu(E\cap F)=-\infty$. To avoid these indeterminate sums we make our restriction.

On the other hand, there are certain pathologies we don’t have to worry about. For instance, if $\{E_n\}$ is a pairwise disjoint sequence of measurable sets, then countable additivity tells us that

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

and so the sum either converges (if the measure of the union is finite) or it definitely diverges to $\pm\infty$. That is, even though we may have negative terms we don’t have to worry about an oscillating sum that fails to converge because the sequence of partial sums jumps around and never settles down. So it always makes sense to write such a sum down, even if its value may be infinite.

All the language about a measure being finite, $\sigma$-finite, totally finite, or totally $\sigma$-finite carries over. The only modification is that we have to ask for $\lvert\mu(E)\rvert<\infty$ or (equivalently) $-\infty<\mu(E)<\infty$ instead of $\mu(E)<\infty$ in the definitions.

Of course, just as for measures, signed measures are finitely additive (which we used above) and thus subtractive. It won’t be monotone, in general, though; Given a set $E$ and a subset $F\subseteq E$ we can write

$\displaystyle\mu(E)=\mu(E\setminus F)+\mu(F)$

If $\mu(E\setminus F)<0$ then $\mu(E)<\mu(F)$, even though $F$ is the subset. However, we can at least say that if $\lvert\mu(E)\rvert<\infty$ then $\lvert\mu(F)\rvert<\infty$ as well. Indeed, using the same equation if either one of the summands on the right is infinite then $\mu(E)$ is as well. If both are, then they’re both infinite in the same direction (since $\mu$ can only assume one of $\pm\infty$) and so $\mu(E)$ is again infinite. The only possibility for a finite $\mu(E)$ is for both $\mu(E\setminus F)$ and $\mu(F)$ to be finite.

June 22, 2010 Posted by | Analysis, Measure Theory | 7 Comments