Jordan Decomposition Theorem

Theorem

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

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

Let $\tuple {P, N}$ be a Hahn decomposition of $\mu$.

Then there exists measures $\mu^+$ and $\mu^-$ on $\struct {X, \Sigma}$ such that:

$\mu = \mu^+ - \mu^-$

Further, at least one of $\mu^+$ and $\mu^-$ is finite.

Proof

From the definition of a Hahn decomposition, the set $P$ is $\mu$-positive, the set $N$ is $\mu$-negative and:

$X = P \cup N$

with $P$ and $N$ disjoint.

From Sigma-Algebra Closed under Countable Intersection, we have:

$A \cap P \in \Sigma$

and:

$A \cap N \in \Sigma$

for each $A \in \Sigma$.

Now, for each $A \in \Sigma$, define:

$\map {\mu^+} A = \map \mu {A \cap P}$

and:

$\map {\mu^-} A = -\map \mu {A \cap N}$

We verify that $\mu^+$ and $\mu^-$ are indeed measures by first showing that they are signed measures.

Lemma

$\mu^+$ and $\mu^-$ are signed measures.

$\Box$

We show that these are in fact measures by showing that $\mu^+ \ge 0$ and $\mu^- \ge 0$.

From Intersection is Subset, we have:

$A \cap P \subseteq P$

Since $P$ is a $\mu$-positive set, we have:

$\map \mu {A \cap P} \ge 0$

for each $A \in \Sigma$.

From Intersection is Subset, we have:

$A \cap N \subseteq N$

Since $N$ is a $\mu$-negative set, we have:

$\map \mu {A \cap N} \le 0$

for each $A \in \Sigma$.

So:

$\map {\mu^+} A \ge 0$

and:

$\map {\mu^-} A \ge 0$

for each $A \in \Sigma$.

So, from Non-Negative Signed Measure is Measure, we have:

$\mu^+$ and $\mu^-$ are measures.

We now verify that at least one of the measures are finite.

From Intersection with Subset is Subset, we have:

$\map {\mu^+} X = \map \mu {X \cap P} = \map \mu P$

and:

$\map {\mu^-} X = -\map \mu {X \cap N} = -\map \mu N$

From the definition of a finite measure, we show that either:

$\map \mu P < \infty$

or:

$-\map \mu N < \infty$

Since $\mu$ is countably additive, we have that:

$\map \mu X = \map \mu P + \map \mu N$

From Set is Subset of Itself, we have:

$P \subseteq P$

so, since $P$ is $\mu$-positive, we have:

$\map \mu P \ge 0$

Applying Set is Subset of Itself again we have:

$N \subseteq N$

so, since $N$ is $\mu$-negative, we have:

$\map \mu N \le 0$

By the definition of a signed measure, we cannot have both $\map \mu P = +\infty$ and $\map \mu N = -\infty$, so at least one of $\map \mu P$ and $\map \mu N$ is finite.

So, either:

$\map {\mu^+} X = \map \mu P < \infty$

or:

$\map {\mu^-} X = -\map \mu N < \infty$

That is, at least one of $\mu^+$ and $\mu^-$ are finite.

Finally, we verify that:

$\mu = \mu^+ - \mu^-$

Note that $A \cap P$ and $A \cap N$ are disjoint with:

$A = \paren {A \cap P} \cup \paren {A \cap N}$

so:

\(\ds \map \mu A\)

\(=\)

\(\ds \map \mu {A \cap P} + \map \mu {A \cap N}\)

since $\mu$ is countably additive

\(\ds \)

\(=\)

\(\ds \map \mu {A \cap P} - \paren {-\map \mu {A \cap N} }\)

\(\ds \)

\(=\)

\(\ds \map {\mu^+} A - \map {\mu^-} A\)

So:

$\mu = \mu^+ - \mu^-$

as required.

$\blacksquare$

Source of Name

This entry was named for Marie Ennemond Camille Jordan.

Sources

• 2013: Donald L. Cohn: Measure Theory (2nd ed.) ... (previous) ... (next): $4.1$