The Unapologetic Mathematician

Mathematics for the interested outsider

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


  1. […] Measures and Sequences We have a couple results about signed measures and certain sequences of […]

    Pingback by Signed Measures and Sequences « The Unapologetic Mathematician | June 23, 2010 | Reply

  2. […] Given a signed measure on a measurable space , we can use it to break up the space into two pieces. One of them will […]

    Pingback by Hahn Decompositions « The Unapologetic Mathematician | June 24, 2010 | Reply

  3. […] It’s not too hard to construct examples showing that Hahn decompositions for a signed measure, though they exist, are not unique. But if we have two of them — — there’s […]

    Pingback by Jordan Decompositions « The Unapologetic Mathematician | June 25, 2010 | Reply

  4. […] — so the whole space is measurable — and we restrict our attention to totally finite signed measures. These form a vector space, since the sum of two finite signed measures is again a finite signed […]

    Pingback by The Banach Space of Totally Finite Signed Measures « The Unapologetic Mathematician | June 28, 2010 | Reply

  5. […] Now, as we’ve pointed out, this will be a measure so long as is a.e. non-negative. But now if is any integrable function at all, the indefinite integral is a finite signed measure. […]

    Pingback by The Jordan Decomposition of an Indefinite Integral « The Unapologetic Mathematician | June 29, 2010 | Reply

  6. […] with Respect to a Signed Measure If is a signed measure then we know that the total variation is a measure. It then makes sense to discuss whether or not […]

    Pingback by Integration with Respect to a Signed Measure « The Unapologetic Mathematician | June 30, 2010 | Reply

  7. […] The Radon-Nikodym Theorem for Signed Measures Now that we’ve proven the Radon-Nikodym theorem, we can extend it to the case where is a -finite signed measures. […]

    Pingback by The Radon-Nikodym Theorem for Signed Measures « The Unapologetic Mathematician | July 8, 2010 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: