Restricted Measure is Measure

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $\Sigma'$ be a sub-$\sigma$-algebra of $\Sigma$.

Then the restricted measure $\mu \restriction_{\Sigma'}$ is a measure on the measurable space $\struct {X, \Sigma'}$.

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

Axiom $(1)$
The statement of axiom $(1)$ for $\mu \restriction_{\Sigma'}$ is:


 * $\forall E' \in \Sigma': \map {\mu \restriction_{\Sigma'} } {E'} \ge 0$

Now, for every $E' \in \Sigma'$, compute:

Axiom $(2)$
Let $\sequence {E'_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma'$.

Then the statement of axiom $(2)$ for $\mu \restriction_{\Sigma'}$ is:


 * $\ds \map {\mu \restriction_{\Sigma'} } {\bigcup_{n \mathop \in \N} E'_n} = \sum_{n \mathop \in \N} \map {\mu \restriction_{\Sigma'} } {E'_n}$

One can show this by means of the following computation:

Axiom $(3')$
The statement of axiom $(3')$ for $\mu \restriction_{\Sigma'}$ is:


 * $\map {\mu \restriction_{\Sigma'} } \O = 0$

By Sigma-Algebra Contains Empty Set, $\O \in \Sigma'$. Hence:


 * $\map {\mu \restriction_{\Sigma'} } \O = \map \mu \O = 0$

because $\mu$ is a measure.

Having verified a suitable set of axioms, it follows that $\mu \restriction_{\Sigma'}$ is a measure.