Hahn Decomposition Theorem

Theorem
Let $\struct {X, \Sigma}$ be a measurable space.

Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.

Then there exists disjoint sets $P$ and $N$ such that:


 * $(1): \quad$ $P$ is a $\mu$-positive set and $N$ is a $\mu$-negative set
 * $(2): \quad$ $X = P \cup N$
 * $(3): \quad$ for any other $\mu$-positive set $P'$ and $\mu$-negative set $N'$ with $X = P' \cup N'$, the symmetric differences $P \Delta P'$ and $N \Delta N'$ are $\mu$-null.

Proof
Note that $\mu$ can attain at most one of $+\infty$ and $-\infty$.

Suppose first that $\mu$ does not attain the value $-\infty$.

Set:


 * $s_1 = \inf \set {\map \mu D : D \in \Sigma \text { and } D \subseteq X}$

Since:


 * $\emptyset \in \set {\map \mu D : D \in \Sigma \text { and } D \subseteq X}$

and $\map \mu \emptyset = 0$, we have $s_1 \le 0$.

From the definition of infimum, we can pick a $\Sigma$-measurable set $D_1 \subseteq X$ such that:


 * $\ds \map \mu {D_1} \le \max \set {\frac {s_1} 2, -1} \le 0$

From Measurable Set with Negative Measure has Negative Subset, there exists a $\mu$-negative set $N_1 \subseteq D_1$ such that:


 * $\map \mu {N_1} \le \map \mu {D_1}$

We define the sequences $\sequence {s_n}_{n \in \N}$ and $\sequence {N_n}_{n \in \N}$ recursively.

We ensure that the sequence $\sequence {N_n}$ is a pairwise disjoint family of sets.

For $n > 1$ set:


 * $\ds s_n = \inf \set {\map \mu D : D \in \Sigma \text { and } D \subseteq X \setminus \paren {\bigcup_{i = 1}^{n - 1} N_i} }$

Since:


 * $\ds \emptyset \in \set {\map \mu D : D \in \Sigma \text { and } D \subseteq X \setminus \paren {\bigcup_{i = 1}^{n - 1} N_i} }$

and $\map \mu \emptyset = 0$, we have $s_n \le 0$.

From the definition of infimum, we can pick a $\Sigma$-measurable set:


 * $\ds D_n \subseteq X \setminus \paren {\bigcup_{i = 1}^{n - 1} N_i}$

such that:


 * $\ds \map \mu {D_n} \le \max \set {\frac {s_n} 2, -1} \le 0$

From Measurable Set with Negative Measure has Negative Subset, there exists a $\mu$-negative set $N_n \subseteq D_n$ such that:


 * $\map \mu {N_n} \le \map \mu {D_n}$

Clearly $N_n$ is disjoint to each of $N_1, N_2, \ldots, N_{n - 1}$.

So the thus constructed $\sequence {N_n}$ is a pairwise disjoint family of sets.

We now prove $(1)$ and $(2)$.

Set:


 * $\ds N = \bigcup_{i \mathop = 1}^\infty N_i$

We show that $N$ is $\mu$-negative.

We have, for every $B \subseteq N$ that:

Since $N_i$ is $\mu$-negative, and:


 * $B \cap N_i \subseteq N_i$

we have:


 * $\map \mu {B \cap N_i} \le 0$

so:


 * $\ds \map \mu B = \sum_{i \mathop = 1}^\infty \map \mu {B \cap N_i} \le 0$

Since $B$ was an arbitrary $\Sigma$-measurable subset of $N$, we have that:


 * $N$ is $\mu$-negative.

Now set:


 * $P = X \setminus N$

Clearly:


 * $X = P \cup N$

It remains to show that $P$ is $\mu$-positive.

suppose that $A \subseteq P$ has $\map \mu A < 0$.

Note that since $P = X \setminus N$, we have:


 * $\ds A \subseteq P \subseteq X \setminus \paren {\bigcup_{i = 1}^{n - 1} N_i}$

for each $n$.

So:


 * $s_n \le \map \mu A < 0$

In particular we cannot have $s_n \to 0$.

We then have:

Each term of the series is negative, and we do not have $\tfrac {s_n} 2 \to 0$, so we have:


 * $\ds \map \mu N = \sum_{i \mathop = 1}^\infty \max \set {\frac {s_n} 2, -1}$

But $\mu$ does not take the value $-\infty$, so we have a contradiction.

So $P$ is $\mu$-positive.

We now prove $(3)$.

We have that:

and:

From Intersection of Positive Set and Negative Set is Null Set, we have:


 * $\map \mu {N \cap P'} = 0$

and:


 * $\map \mu {N' \cap P} = 0$

From Null Sets Closed under Countable Union, we then have:


 * $\map \mu {P \Delta P'} = \map \mu {N \Delta N'} = 0$

proving $(3)$.