Characterization of Null Sets of Variation of Signed Measure

Theorem

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

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

Let $\size \mu$ be the variation of $\mu$.

Then $A \in \Sigma$ is such that $\map {\size \mu} A = 0$ if and only if:

for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$.

Proof

Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.

Then, from the definition of variation:

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

Suppose that $A \in \Sigma$ is such that $\map {\size \mu} A = 0$.

Then:

$\map {\mu^+} A + \map {\mu^-} A = 0$

Since $\mu^+ \ge 0$ and $\mu^- \ge 0$ we have:

$\map {\mu^+} A = 0 $ and $\map {\mu^-} A = 0$

From Null Sets Closed under Subset, we then have:

for each $\Sigma$-measurable set $B \subseteq A$, we have $\map {\mu^+} B = 0$

and:

for each $\Sigma$-measurable set $B \subseteq A$, we have $\map {\mu^-} B = 0$

So, for each $\Sigma$-measurable set $B \subseteq A$, we have:

$\map {\mu^+} B - \map {\mu^-} B = 0$

From the definition of the Jordan decomposition, we therefore have:

$\map \mu B = 0$

for each $\Sigma$-measurable set $B \subseteq A$.

$\Box$

Now suppose that:

for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$.

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

From the definition of Jordan decomposition, and Uniqueness of Jordan Decomposition, we have:

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

and:

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

From Intersection is Subset, we have:

$P \cap A \subseteq A$

and:

$N \cap A \subseteq A$

So, we have:

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

and:

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

from the assumption.

That is:

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

and:

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

So:

$\map {\size \mu} A = \map {\mu^+} A + \map {\mu^-} A = 0$

So, if:

for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$

then:

$\map {\size \mu} A = 0$

$\blacksquare$