I try to solve the following task: Let $\lambda_n$ be the Lebesgue measure in $\mathbb{R}^n$. Consider an arbitrary subset $A \in M_{\lambda_n^*}$ such that $\lambda_n(A)<\infty$.
Show that there are open sets $(U_n)_{n\geq 1}$ such that $$\lambda_n(A)=\lambda_n\left(\bigcap_{n\geq 1}U_n\right)$$ In particular, the Borel set $B=\bigcap_{n\geq 1}U_n$ is such that $A \subset B$ and $\lambda_n(A)=\lambda_n(B)$.
This is what I've thought so far:
let $B = \bigcap_{n\geq 1}U_n$ such that $A\subset B$, then from monotonie of $\lambda_n$ it follows that $$\lambda_n(A) \leq \lambda_n(B)$$
Now we have to show $$\lambda_n(A) \geq \lambda_n(B)$$
So if we can write the intersection $B$ as the union of open sets, we could define an $\varepsilon > 0$ and from the definition of the Lebesgue outer measure there exist a sequence of open n-dimensional interval $\{R_i\}_{i\in\mathbb{N}}$ with $A \subset \bigcup_{n\geq 1}R_i$ and $$\lambda_n(A) \geq \sum\limits_{i=1}^n vol(R_i) -\varepsilon$$ then from subadditivity of $$\lambda_n(\bigcup_{n\geq 1} R_i) \leq \sum\limits_{i=1}^n \lambda_n(R_i) = \sum\limits_{i=1}^n vol(R_i)\leq \lambda_n(A) + \varepsilon$$ And because we can chose $\varepsilon$, we are done.
Now is this correct so far? And how could I write the intersection $B$ as a union of open sets?