Restricting Measure Preserves Finiteness

Theorem
Let $\left({X, \mathcal A, \mu}\right)$ be a measure space.

Let $\mu$ be a finite measure.

Let $\mathcal B$ be a sub-$\sigma$-algebra of $\mathcal A$.

Then the restricted measure $\mu \restriction_{\mathcal B}$ is also a finite measure.

Proof
By Restricted Measure is Measure, $\mu \restriction_{\mathcal B}$ is a measure.

Now by definition of $\mu \restriction_{\mathcal B}$, have:


 * $\mu \restriction_{\mathcal B} \left({X}\right) = \mu \left({X}\right) < \infty$

as $\mu$ is a finite measure.

Hence $\mu \restriction_{\mathcal B}$ is also a finite measure.