Restricted Measure is Measure

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

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

Then the restricted measure $\mu \restriction_{\mathcal B}$ is a measure on the measurable space $\left({X, \mathcal B}\right)$.

Proof
Verify the axioms for a measure in turn for $\mu \restriction_{\mathcal B}$:

$(1)$
For every $B \in \mathcal B$ have:

$(2)$
For every sequence $\left({B_n}\right)_{n \in \N}$ in $\mathcal B$, have:

$(3')$
By Sigma-Algebra Contains Empty Set, $\varnothing \in \mathcal B$. Hence:


 * $\mu \restriction_{\mathcal B} \left({\varnothing}\right) = \mu \left({\varnothing}\right) = 0$

because $\mu$ is a measure.

Having verified explicitly the conditions, conclude that $\mu \restriction_{\mathcal B}$ is a measure.