Restricting Measure Preserves Finiteness
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Theorem
Let $\left({X, \Sigma, \mu}\right)$ be a measure space.
Let $\mu$ be a finite measure.
Let $\Sigma'$ be a sub-$\sigma$-algebra of $\Sigma$.
Then the restricted measure $\mu \restriction_{\Sigma'}$ is also a finite measure.
Proof
By Restricted Measure is Measure, $\mu \restriction_{\Sigma'}$ is a measure.
Now by definition of $\mu \restriction_{\Sigma'}$, have:
- $\mu \restriction_{\Sigma'} \left({X}\right) = \mu \left({X}\right) < \infty$
as $\mu$ is a finite measure.
Hence $\mu \restriction_{\Sigma'}$ is also a finite measure.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 4$: Problem $14 \ \text{(ii)}$