# 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:

\(\ds \map \mu B\) | \(=\) | \(\ds \map \mu {B \cap N}\) | Intersection with Subset is Subset | |||||||||||

\(\ds \) | \(=\) | \(\ds \map \mu {\bigcup_{i \mathop = 1}^\infty \paren {B \cap N_i} }\) | Intersection Distributes over Union of Family | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^\infty \map \mu {B \cap N_i}\) | $\mu$ is countably additive |

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.

Aiming for a contradiction, suppose 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:

\(\ds \map \mu N\) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \map \mu {N_n}\) | $\mu$ is countably additive | |||||||||||

\(\ds \) | \(\le\) | \(\ds \sum_{n \mathop = 1}^\infty \map \mu {D_n}\) | since $\map \mu {N_n} \le \map \mu {D_n}$ for each $i$ | |||||||||||

\(\ds \) | \(\le\) | \(\ds \sum_{n \mathop = 1}^\infty \max \set {\frac {s_n} 2, -1}\) |

Aiming for a contradiction, suppose, 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:

\(\ds P \Delta P'\) | \(=\) | \(\ds \paren {P \setminus P'} \cup \paren {P' \setminus P}\) | Definition of Symmetric Difference | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {P \cap \paren {X \setminus P'} } \cup \paren {P' \cap \paren {X \setminus P} }\) | Definition of Set Difference | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {P \cap N'} \cup \paren {P' \cap N}\) | since $X \setminus P' = N$ and $X \setminus P = N$ |

and:

\(\ds N \Delta N'\) | \(=\) | \(\ds \paren {N \setminus N'} \cup \paren {N' \setminus N}\) | Definition of Symmetric Difference | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {N \cap \paren {X \setminus N'} } \cup \paren {N' \cap \paren {X \setminus N} }\) | Definition of Set Difference | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {N \cap P'} \cup \paren {N' \cap P}\) | since $X \setminus N' = P'$ and $X \setminus N = P$ | |||||||||||

\(\ds \) | \(=\) | \(\ds P \Delta P'\) |

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 if and only if it is $\mu$-negative, and $\nu$-negative if and only if it is $\mu$-positive.

Note that a set $A \in \Sigma$ is $\nu$-positive if and only if 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 if and only if it is $\mu$-negative.

Now note that a set $A \in \Sigma$ is $\nu$-negative if and only if 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 if and only if 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 if and only if it is $\mu$-null.

Note that $A \in \Sigma$ is $\nu$-null if and only if 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 if and only if 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.

$\blacksquare$

## Source of Name

This entry was named for Hans Hahn.

## Sources

- 2013: Donald L. Cohn:
*Measure Theory*(2nd ed.) ... (previous) ... (next): $4.1$: Signed and Complex Measures