Signed Measure may not be Monotone

Theorem

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

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

Then $\mu$ may not be monotone.

Proof

Let:

$\struct {X, \Sigma} = \struct {\R, \map \BB \R}$

where $\map \BB \R$ is the Borel $\sigma$-algebra on $\R$.

Define:

$\mu = \delta_1 - 2 \delta_2$

where $\delta_1$ and $\delta_2$ are the Dirac measures at $1$ and $2$ respectively.

Since $\delta_1$ and $\delta_2$ are both finite measures, we have:

$\mu$ is a signed measure

from Linear Combination of Signed Measures is Signed Measure.

Then, we have:

$\closedint 0 1 \subseteq \closedint 0 2$

with:

\(\ds \map \mu {\closedint 0 1}\)

\(=\)

\(\ds \map {\delta_1} {\closedint 0 1} - 2 \map {\delta_2} {\closedint 0 1}\)

\(\ds \)

\(=\)

\(\ds 1 - 0\)

Definition of Dirac Measure

\(\ds \)

\(=\)

\(\ds 1\)

and:

\(\ds \map \mu {\closedint 0 2}\)

\(=\)

\(\ds \map {\delta_1} {\closedint 0 2} - 2 \map {\delta_2} {\closedint 0 2}\)

\(\ds \)

\(=\)

\(\ds 1 - 2\)

Definition of Dirac Measure

\(\ds \)

\(=\)

\(\ds -1\)

So:

$\closedint 0 1 \subseteq \closedint 0 2$ and $\map \mu {\closedint 0 2} \le \map \mu {\closedint 0 1}$

So:

$\mu$ is not monotone.

$\blacksquare$