# 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.

$\blacksquare$