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?

  • $\begingroup$ You forgotto require $A\subseteq U_n$, I suppose. $\endgroup$ – Hagen von Eitzen Nov 1 '15 at 14:01
  • $\begingroup$ Well this is the point. I have $B = \bigcap_{n\in\mathbb{N}} U_n$ and $A \subset B$ but I don't know how to write this as a union of open sets. $\endgroup$ – Matriz Nov 1 '15 at 15:59

Let $A\in B(R^n)$ sucht that $\lambda_n (A)<\infty$, let $U_m:=\{x\in R^n:d(x,A)<1/m\}$ it is possible to show that

  1. $\lambda_n (U_m)<\infty$
  2. $U_m$ is open
  3. $A=\cap_{m\geq 1}U_m$

By the dominated convergence theorem using indicator functions the result follows.


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