Prove that $\mu^*$ restricted to $\mathcal{B}(X)$ is a measure on $\mathcal{B}(X)$.

Question

We have the following theorem:

If $\mu^*$ is a metric outer measure, then all Borel sets in $X$ are $\mu^*$-measureable.

I have found the proof in here, but I can't see clearly everything so I need some help. I don't understand the part when we want to show that $\lim_{n\rightarrow\infty} \nu(E_n)=\nu(E\cap F').$ It's on the 2nd page. Could you explain me why the inequalities below are satisfied ((4) and the one above it)? Any tips will be extreamely helpful! Maybe you know other easy proofs of the theorem you can share? Thanks in advance!

Answer 1

• $\rho(y, F) \leq \rho(x, y) + \rho(x, F) < \frac{1}{n(n + 1)}+\frac{1}{n + 1}=\frac{1}{n}$:

It's just a matter of choice of $x$ and $y$. The first inequality is triangle inequality. Now, $x\in D_{n+1}$ which means (by definition of $D_{n+1}$ ) that $x \notin E_{n+1}$; in other words (by definition of $E_{n+1}$) $\rho(x,F) < \frac{1}{n+1}$. Furthermore, $y\in X$ was chosen such that $\rho (x,y)< \frac{1}{n(n+1)}$.

• $\rho(D_{n+1}, E_n)\geq \frac{1}{n(n+1)}$ :

Lets assume this not true, i.e. $\rho(D_{n+1}, E_n)< \frac{1}{n(n+1)}$. Then we can find a $x\in D_{n+1}$ and an $y\in E_n$ such that $\rho(x,y)< \frac{1}{n(n+1)}$, and by the previous inequality, this means that
$\rho(y, F)< \frac{1}{n}$ which is a contradiction since we assumed that $y$ is an element of $E_n$.