Extensions and Restrictions

Let $(\Omega , \mathcal{A}, \mu )$ denote a measure space. Pick $E \in \mathcal{A}$ and define $\mathcal{A}_{E}=\left\{ F \in \mathcal{A}: F \subseteq E\right\}$. Then $\mathcal{A}_{E}$ is a $\sigma$-field on $E, \mathcal{A}_{E}=\left\{ A \cap E: A \in \mathcal{A}\right\}$, and the restriction $\mu_{E}$ of $\mu$ [from $\mathcal{A}$ ] to $\mathcal{A}_{E}$ is a measure. That is, $\left(E, \mathcal{A}_{E}, \mu_{E}\right)$ is a measure space and $\mu_{E}=\mu$ on $\mathcal{A}_{E}$.

Prove Claim 1 and 2.

Claim 1 Assume the notation of the definition. If $\mu$ is $\sigma$-finite on $\mathscr {G}$, then $v$ is $\sigma$-finite on $\mathscr {G}$ as well.

Claim 2 Suppose that $\varnothing \in \mathscr {G}$, and let $\mathscr {G} \subseteq \mathscr {H} \subseteq 2^{\Omega }$. Let $v: \mathscr {H} \rightarrow \overline{\mathbb {R}}$ denote a measure. Then the restriction of $v$ to $\mathscr {G}$ is a measure.