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 $\O \subseteq X$ and $\map \mu \O = 0$ we have:


 * $0 \in \set {\map \mu D : D \in \Sigma \text { and } D \subseteq X}$

From the definition of infimum, we have $s_1 \le 0$.

Applying the definition of infimum again, we can pick a $D_1 \in \Sigma$ 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 now 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:


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

Again we have:


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

Since $\map \mu \O = 0$, we have:


 * $\ds 0 \in \set {\map \mu D : D \in \Sigma \text { and } D \subseteq X \setminus \paren {\bigcup_{i = 1}^{n - 1} N_i} }$

and so from the definition of infimum we have $s_n \le 0$.

Applying the definition of infimum again, 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:


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

Since:


 * $\ds N_n \not \in \bigcup_{i \mathop = 1}^{n - 1} N_i$

we have that $N_n$ 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, from countable additivity:


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

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:


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

for each $n$.

So:


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

If $\sequence {s_n}_{n \in \N}$ converges, then:


 * $\ds \lim_{n \mathop \to \infty} s_n \le \map \mu A < 0$

from Lower and Upper Bounds for Sequences.

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

We have:

, suppose that:


 * $\ds \sum_{n \mathop = 1}^\infty \max \set {\frac {s_n} 2, -1}$ converges.

Then from Terms in Convergent Series Converge to Zero, we have:


 * $\ds \max \set {\frac {s_n} 2, -1} \to 0$

In particular, there exists $N$ such that:


 * $\ds \size {\max \set {\frac {s_n} 2, -1} } < 1$

for $n > N$.

That is:


 * $\ds \max \set {\frac {s_n} 2, -1} = \frac {s_n} 2$

for $n > N$.

So if:


 * $\ds \max \set {\frac {s_n} 2, -1} \to 0$

we have:


 * $\dfrac {s_n} 2 \to 0$

That is, from Multiple Rule for Real Sequences:


 * $s_n \to 0$

We have established that this is impossible, so we have reached a contradiction, and:


 * $\ds \sum_{n \mathop = 1}^\infty \max \set {\frac {s_n} 2, -1}$ does not converge.

Since:


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

for each $n$, we have:


 * $\ds \sum_{n \mathop = 1}^\infty \max \set {\frac {s_n} 2, -1} = -\infty$

So:


 * $\map \mu N \le -\infty$

that is:


 * $\map \mu N = -\infty$

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:


 * $N \cap P'$ is a $\mu$-null set

and:


 * $N' \cap P$ is a $\mu$-null set.

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


 * $P \Delta P' = N \Delta N'$ is $\mu$-null

proving $(3)$.

Now suppose that $\mu$ does not attain the value $+\infty$.

Then the signed measure $\nu = -\mu$ does not attain the value $-\infty$.

So, we can find disjoint sets $P$ and $N$ such that:


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

We show that this yields an appropriate decomposition for $\mu$.

We first show that a set $A \in \Sigma$ is $\nu$-positive it is $\mu$-negative, and $\nu$-negative  it is $\mu$-positive.

Note that a set $A \in \Sigma$ is $\nu$-positive for all $B \in \Sigma$ with $B \subseteq A$ we have:


 * $\ds \map \nu B = \map {-\mu} B \ge 0$

This is equivalent to:


 * $\map \mu B \le 0$

So $A$ is $\nu$-positive it is $\mu$-negative.

Now note that a set $A \in \Sigma$ is $\nu$-negative for all $B \in \Sigma$ with $B \subseteq A$ we have:


 * $\ds \map \nu B = \map {-\mu} B \le 0$

This is equivalent to:


 * $\map \mu B \ge 0$

So $A$ is $\nu$-negative it is $\mu$-positive. Note that if a set $A \in \Sigma$ is $\nu$-null, we have that:


 * $-\map \mu A = 0$

so:


 * $\map \mu A = 0$

That is, $P$ is $\mu$-negative and $N$ is $\mu$-positive.

At this point it is convenient to relabel $\PP = N$ and $\NN = N$.

$(2')$ requires no adaption, so we move to $(3')$.

We show that $A \in \Sigma$ is $\nu$-null it is $\mu$-null.

Note that $A \in \Sigma$ is $\nu$-null for all $B \in \Sigma$ with $B \subseteq A$, we have:


 * $\map \nu B = 0$

This is equivalent to:


 * $\map {-\mu} B = 0$

that is:


 * $\map \mu B = 0$

So $A \in \Sigma$ is $\nu$-null it is $\mu$-null.

Relabelling $\PP' = N'$ and $\NN' = P'$, we find that $(3')$ is equivalent to:


 * for any other $\mu$-negative set $\NN'$ and $\mu$-positive set $\PP'$ with $X = \NN' \cup \PP'$, the symmetric differences $\PP \Delta \PP'$ and $\NN \Delta \NN'$ are $\mu$-null.

So we have:


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

in the case that $\mu$ does not take the value $+\infty$.

We therefore have the demand in both cases.