\begin{proof} of 1: First, using Lebesgue Outer Measure we have that $$\mu^{*}(E)=\inf\left\{ \sum_{j=1}^{\infty}\lambda(A_{j}):A(j)_{j\ge 1}\subseteq \mathscr{A}\text{ s.t. }E\subseteq \bigcup_{j=1}^{\infty}A_{j} \right\}$$is an Outer measure. Then using we get that $$\mathscr{M}=\{ A\subseteq X:A\text{ is }\mu^{*}\text{-measurable} \}$$is a σ-algebra and $(X,\mathscr{M},\left.\mu^{*}\right|_{\mathscr{M}})$ is a Complete Measure Space. Then using Construction of Outer Measure from Pre-measure we get that $\left.\mu^{*}\right|_{\mathscr{A}}=\lambda$. Then since $\mathscr{A}\subseteq \mathscr{M}$ from the second term of that proposition we get that $\mathcal{B}$, which is the Smallest σ-algebra generated by $\mathscr{A}$ satisfies the inclusion $$\mathscr{A}\subseteq \mathcal{B}\subseteq \mathscr{M}$$and since $\mathcal{B}\subseteq \mathscr{M}$ are σ-algebras. then $\left.\mu^{*}\right|_{\mathcal{B}}=:\mu$ is a Measure on $\mathcal{B}$. Then finally since we have $\mathscr{A}\subseteq \mathcal{B}$ we have that $\left.\mu\right|_{\mathscr{A}}=:\lambda$. \end{proof} >[!thm] (wikipedia) >Let $I$ be a semiring of sets on $X$ and let $\mu:R\to[0,\infty]$ be a pre-measure on $R$: >1. Then $\mu$ has an extension (i.e. a measure $\tilde{\mu}$) $$\tilde{\mu}:\sigma(R)\to[0,\infty]$$ >2. If $X$ is σ-finite, then $\tilde{\mu}$ is unique, and $\tilde{\mu}$ is also σ-finite.