Characterization of Null Sets of Variation of Complex Measure

Theorem

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

Let $\mu$ be a complex 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 $A \in \Sigma$.

Let $\map P A$ be the set of finite partitions of $A$ into $\Sigma$-measurable sets.

From the definition of variation, we have:

$\ds \map {\cmod \mu} A = \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$

Suppose that:

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

Then, for any:

$\set {A_1, A_2, \ldots, A_n} \in \map P A$

we have:

$\map \mu {A_i} = 0$ for each $i$

so:

$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} } = 0$

This gives:

$\ds \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} = \set 0$

So, from the definition of supremum, we have:

$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} = 0$

so:

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

$\Box$

Suppose that:

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

Then:

$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} = 0$

Note that since each element of:

$\ds \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$

is non-negative, we must have:

$\ds \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} = \set 0$

So, for all:

$\set {A_1, A_2, \ldots, A_n} \in \map P A$

we have:

$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} } = 0$

Let $B \subseteq A$ be $\Sigma$-measurable.

From Sigma-Algebra Closed under Set Difference, we have:

$A \setminus B$ is $\Sigma$-measurable.

We also have:

$A = B \cup \paren {A \setminus B}$

so:

$\set {B, A \setminus B}$ is a finite partition of $A$ into $\Sigma$-measurable sets.

That is:

$\set {B, A \setminus B} \in \map P A$

We therefore have:

$\cmod {\map \mu B} + \cmod {\map \mu {A \setminus B} } = 0$

Since:

$\cmod {\map \mu {A \setminus B} } \ge 0$

and:

$\cmod {\map \mu B} \ge 0$

we must have:

$\cmod {\map \mu B} = \cmod {\map \mu {A \setminus B} } = 0$

and in particular:

$\map \mu B = 0$

So:

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

$\blacksquare$