Signed Measure of Limit of Decreasing Sequence of Measurable Sets

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

Let $E \in \Sigma$.

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


 * $E_n \downarrow E$

where $E_n \downarrow E$ denotes limit of decreasing sequence of sets, and:


 * there exists $m \in \N$ such that $\map \mu {E_m}$ is finite.

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:


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

and:


 * $\map \mu {E_m} = \map {\mu^+} {E_m} - \map {\mu^-} {E_m}$

Since $\map \mu {E_m}$ is finite, we have that:


 * $\map {\mu^+} {E_m}$ and $\map {\mu^-} {E_m}$ are finite.

From Measure of Limit of Decreasing Sequence of Measurable Sets: Corollary, 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:

Also see

 * Measure of Limit of Decreasing Sequence of Measurable Sets