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.

June 24, 2010 Posted by | Analysis, Measure Theory | 6 Comments

   

Follow

Get every new post delivered to your Inbox.

Join 366 other followers