Royden 17.1.6 p. 341

\textbf{Exercise 17.1.6} Let $(X,\mathcal{M},\mu)$ be a measure space and let $X_0\in{\mathcal{M}}$. Define $\mathcal{M}_0$ to be the collection of sets in $\mathcal{M}$ that are subsets of $X_0$ and $\mu_0 = \mu|_{X_0}$. Show that $(X_0,\mathcal{M}_0,\mu_0)$ is a measure space.

This problem establishes that a "smaller" measure space can be defined within a measurable set.

\textit{Proof} : To show that $(X_0,\mathcal{M}_0,\mu_0)$ is a measure space, we need to show
(i) $\mathcal{M}_0 \neq \emptyset$
(ii) $\mathcal{M}_0$ is a $\sigma$-algebra (closed under compliments and countable unions and intersections)
(iii) $\mu_0 = \mu|_{X_0}$ is a measure.

(i) $\emptyset \in{\mathcal{M}}$ since $\mu(\emptyset)=0$. Also, $\emptyset \subseteq X_0$ since the empty set is always a subset of any set. Thus, $\emptyset$ is a measurable subset of $X_0$, thus $\emptyset \in{\mathcal{M}_0}$. So, $\mathcal{M}_0$ contains at least one set (the empty set), so is nonempty.

(ii) Let $E\in{\mathcal{M}_0}$. This means that $E\in{\mathcal{M}}$ and $E \subseteq X_0$. Since $\mathcal{M}$ is a $\sigma$-algebra, $E^c\in{\mathcal{M}}$. So, $E^c \cap X_0 \in{\mathcal{M}}$ (an intersection of 2 measurable sets). Also, $E^c \cap X_0$ is the relative compliment of $E$ in $X_0$. If we define our compliments in $\mathcal{M}_0$ in this way, then we get that the relative compliment of $E$ in $X_0$ is measurable and contained in $X_0$, thus in $\mathcal{M}_0$. This makes $\mathcal{M}_0$ closed under (relative) compliments in $X_0$. To show that $\mathcal{M}_0$ is closed under countable set unions, let $\{E_j\}_{j=1}^\infty$ be a sequence of sets s.t. $E_j \in{\mathcal{M}}$ and $E_j \subseteq X_0$ $\forall j\in{\mathbb{N}}$. $E_j \subseteq X_0 \quad \forall j \Longrightarrow \cup_{j=1}^\infty E_j \subseteq X_0$. $\mathcal{M}$ is a $\sigma$-algebra, so $\cup_{j=1}^\infty E_j \subseteq \mathcal{M}$. Thus, $\cup_{j=1}^\infty E_j \in{\mathcal{M}_0}$. DeMorgan's laws then automatically make $\mathcal{M}_0$ also closed under countable intersections. So, $\mathcal{M}_0$ is a $\sigma$-algebra.

(iii) $\mu$ is a measure on all of $\mathcal{M}$, so it is automatically a measure on any subset of $\mathcal{M}$. So, the restriction of $\mu$ to $\mathcal{M}_0$ is a measure.
$\therefore \quad (X_0,\mathcal{M}_0, \mu_0)$ is a measure space. $\blacksquare$