# The Unapologetic Mathematician

## 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.

Posted by John Armstrong | Analysis, Measure Theory | 6 Comments

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