Question

If u be a borel measure on a metric space X and suppose there exists a sequence Gn of open sets, X=UnGn and for all n , u(Gn) < Infinity then u is regular.

Answer 1

Let M= collection of all borel measurable subsets $Y$ of X such that

$\mu(Y)=sup\{\mu(K)|K\subset X,K \ closed\}=inf\{\mu(U)|U\subset X,U \ open\}$. We will show M is $\sigma-$ algebra containing all closed sets and hence all open sets. This will show $\mu$ is regular.

Now let $Y\in M.$ Then, for $\epsilon >0,$ there is $U$ open and $K$ closed such that, $\mu(U)<\mu(Y)+\epsilon$$\mu(K)>\mu(Y)-\epsilon$. Hence $\mu(Y^c)> \mu(U^c)+\epsilon$ and $\mu(Y^c)< \mu(K^c)-\epsilon$ Hence M is closed under complementation. Now let $Y_i$ be a countable collection. Then there exists $U_i$ open and $K_{i}$ closed such that $\mu(K_i)>\mu(Y_i)-\epsilon /2^i$ and $\mu(U_i)<\mu(Y_i)+\epsilon /2^i$ Let $K=\cap K_i$ and $U=\cup U_i$ . Then since its given the measure is $\sigma$ finite, $\mu(U)=lim_{n\rightarrow \infty}\mu(U_i)< \mu(Y)+\Sigma_{i=1}^{\infty} \ \epsilon/2^i$ $=\mu(Y)+\epsilon$. Similarly $\mu(K)>\mu(Y)-\epsilon.$ Hence M is closed under countable union. Now we have to show that it contains all closed sets. Let $Y$ be closed. Let $U_n=\{x\in X| d(y,x)<1/n \ \textrm{for some} \ y \in Y \}.$ Then $U_n$ open and $U_1\supset U_2\supset U_3\cdots$ Also as $Y$ is closed, so for any $x\notin Y$ $d(x,Y)=m>0. \Rightarrow m>1/n$ for some $n.$ Hence $\cap_i U_i =Y$ Hence $\mu(Y)= lim \ \mu(U_n)=$ inf $\mu(U_n)$ $\geq \textrm{inf} \ \mu(U) \ \forall \ U\supset Y.$ . But RHS $\geq \mu(Y)$ and hence the equality. For the other side we note $Y$is a closed set and hence is the supremum. Hence all closed sets are in M. Hence done.