The Unapologetic Mathematician

Mathematics for the interested outsider

Hahn Decompositions

Given a signed measure \mu on a measurable space (X,\mathcal{S}), we can use it to break up the space into two pieces. One of them will contribute positive measure, while the other will contribute negative measure. First: some preliminary definitions.

We call a set E\subseteq X “positive” (with respect to \mu) if for every measurable F\in\mathcal{S} the intersection E\cap F is measurable, and \mu(E\cap F)\geq0. That is, it’s not just that E has positive measure, but every measurable part of E has positive measure. Similarly, we say that E is “negative” if for every measurable F the intersection E\cap F is measurable, and \mu(E\cap F)\leq0. For example, the empty set is both positive and negative. It should be clear from these definitions that the difference of two negative sets is negative, and any disjoint countable union of negative sets is negative, and (thus) any countable union at all of negative sets is negative.

Now, for every signed measure \mu there is a “Hahn decomposition” of X. That is, there are two disjoint sets A and B, with A positive and B negative with respect to \mu, and whose union is all of X. We’ll assume that -\infty<\mu(E)\leq\infty , but if \mu takes the value -\infty (and not \infty) the modifications aren’t difficult.

We write \beta=\inf\mu(B), taking the infimum over all measurable negative sets B. We must be able to find a sequence \{B_i\} of measurable negative sets so that the limit of the \mu(B_i) is \beta — just pick B_i so that \beta\leq\mu(B_i)<\beta+\frac{1}{i} — and we can pick the sequence to be monotonic, with B_i\subseteq B_{i+1}. If we define B as the union — the limit — of this sequence, then we must have \mu(B)=\beta. The measurable negative set B has minimal measure \mu(B).

Now we pick A=X\setminus B, and we must show that A is positive. If it wasn’t, there would be a measurable subset E_0\subseteq A with \mu(E_0)<0. This E_0 cannot itself be negative, or else B\uplus E_0 would be negative and we’d have \mu(B\uplus E_0)=\mu(B)+\mu(E_0)<\mu(B), contradicting the minimality of \mu(B).

So E_0 must contain some subsets of positive measure. We let k_1 be the smallest positive integer so that E_0 contains a subset E_1\subseteq E_0 with \mu(E_1)\geq\frac{1}{k_1}. Then observe that

\displaystyle\mu(E_0\setminus E_1)=\mu(E_0)-\mu(E_1)\leq\mu(E_0)-\frac{1}{k_1}<0

So everything we just said about E_0 holds as well for E_0\setminus E_1. We let k_2 be the smallest positive integer so that E_0\setminus E_1 contains a subset E_2\subseteq E_0\setminus E_1 with \mu(E_2)\geq\frac{1}{k_2}. And so on we go until in the limit we’re left with

\displaystyle F_0=E_0\setminus\biguplus\limits_{i=1}^\infty E_i

after taking out all the sets E_i.

Since -\infty<\mu(E_0)<0, the measure of E_0 is finite, and so the measure of any subset of E_0 must be finite as well. Thus the limits of the \frac{1}{k_n} must be zero, so that the measure of the countable disjoint union of all the E_n can converge. And so any remaining measurable set F that can fit into F_0 must have \mu(F)\leq0. That is, F_0 must be a measurable negative set disjoint from B. But we must have

\displaystyle\mu(F_0)=\mu(E_0)-\sum\limits_{i=1}^\infty\mu(E_i)\leq\mu(E_0)<0

which contradicts the minimality of \mu(B) just like E_0 would have if it had been a negative set. And thus the assumption that \mu(E_0)<0 is untenable, and so every measurable subset of A has positive measure.

About these ads

June 24, 2010 - Posted by | Analysis, Measure Theory

6 Comments »

  1. […] 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 — […]

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

  2. […] that implies that is the zero measure. The triangle inequality takes a bit more work. We take a Hahn decomposition for and […]

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

  3. […] give a Hahn decomposition corresponding to . I say that if we […]

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

  4. […] we take the supremum over all measurable functions with everywhere. Indeed, if we take a Hahn decomposition for , then since is measurable so are and . If we take and , then we […]

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

  5. […] let be a Hahn decomposition for . We find that is a -finite measure on , while is a -finite measure on […]

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

  6. […] holds for itself, and so we may assume that is also a measure. We can further simplify by using Hahn decompositions with respect to both and , passing to subspaces on which each of our signed measures has a […]

    Pingback by The Radon-Nikodym Chain Rule « The Unapologetic Mathematician | July 12, 2010 | Reply


Leave a Reply

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

WordPress.com Logo

You are commenting using your WordPress.com 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 )

Google+ photo

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

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 392 other followers

%d bloggers like this: