Signed Measure of Limit of Increasing Sequence of Measurable Sets

Theorem

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

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

Let $E \in \Sigma$.

Let $\sequence {E_n}_{n \mathop \in \N}$ be an increasing sequence of $\Sigma$-measurable sets such that:

$E_n \uparrow E$

where $E_n \uparrow E$ denotes the limit of an increasing sequence of sets.

Then:

$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$

Proof

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

Then $\mu^+$ and $\mu^-$ are measures with:

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

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

Then we have:

$\map \mu E = \map {\mu^+} E - \map {\mu^-} E$

From Measure of Limit of Increasing Sequence of Measurable Sets, we have:

$\ds \map {\mu^+} E = \lim_{n \mathop \to \infty} \map {\mu^+} {E_n}$

and:

$\ds \map {\mu^-} E = \lim_{n \mathop \to \infty} \map {\mu^-} {E_n}$

with at most one of these limits infinite.

So:

\(\ds \map \mu E\)

\(=\)

\(\ds \lim_{n \mathop \to \infty} \map {\mu^+} {E_n} - \lim_{n \mathop \to \infty} \map {\mu^-} {E_n}\)

\(\ds \)

\(=\)

\(\ds \lim_{n \mathop \to \infty} \paren {\map {\mu^+} {E_n} - \map {\mu^-} {E_n} }\)

Difference Rule for Real Sequences

\(\ds \)

\(=\)

\(\ds \lim_{n \mathop \to \infty} \map \mu {E_n}\)

$\blacksquare$