Signed Measure of Limit of Decreasing Sequence of Measurable Sets
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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:
\(\ds \map \mu E\) | \(=\) | \(\ds \map {\mu^+} E - \map {\mu^-} E\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \map {\mu^+} {E_n} - \lim_{n \mathop \to \infty} \map {\mu^-} {E_n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\map {\mu^+} {E_n} - \map {\mu^-} {E_n} }\) | Difference Rule for Real Sequences | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \map \mu {E_n}\) |
$\blacksquare$