Restricting Measure Preserves Finiteness

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 $14 \ \text{(ii)}$