Let $\mathcal{A}$ be a sigma-algebra on $X$ and $\lambda: \mathcal{A} \to [0, \infty]$ a measure on $X$.
Define $\mu: \mathcal{P}(X)\to[0, \infty]$ with $\mu(E):=$inf$\lbrace \lambda(U):U \in \mathcal{A}, E \subset U\rbrace$.
How to show that $\mu$ is an outer measure and for all sets $E \in \mathcal{A}$ it's $\mu(E)=\lambda(E)$?
I got the first and second property. For the third I have:
If $\varepsilon > 0$, then $\exists U \in \mathcal{A}$ with $\lambda(U_i) \leq \mu(E_i)+\frac{1}{2^i}\varepsilon$.
So, $\mu(\bigcup E_i)\leq \sum \mu(E_i)+\varepsilon$.
I don't see how to conclude that $\mu(E)=\lambda(E)$ for all $E \in \mathcal{A}$.
If $E \in \mathcal A$ then $\mu(E) \leq \lambda (E)$ (taking $U=E$ in the definition). On the other hand, for any $U \in \mathcal A$ with $E\subset U$ we have $\lambda (E ) \leq \lambda (U)$ so $\lambda (E ) \leq \inf \{ \lambda (U): U \in \mathcal A, E\subset U\}$ so $\lambda (E ) \leq \mu (E)$.