Proposition: Let $\mu$ be a borel measure such that $\mu(\Bbb{R^d})<+\infty$ Then $\mu$ is regular, i.e $\mu(A)=\inf\{\mu(G)|G \supset A, \text{ G open}\}=\sup\{\mu(K)|K \subseteq A, \text{ K compact}\}$
Using this ,i have to prove that if a Borel measure on $\Bbb{R^d}$ has the property that $\mu(K)< +\infty$ for every compact subset $K$ then ti is regular.
In my attempt i put $\mu_n(A)=\mu(A \cap B(0,n))$ where $B(0,n)$ is the closed ball with radious $n$.
$\mu_n$ satisfies the above proposition and also $\sup_n \mu_n= \mu$.
So i managed to prove the one equality.
To prove the second i was given the hint to put $\mu_n(A)=\mu(A \cap Q_n)$ where $Q_n$ is a dyadic cube, and $\Bbb{R^d}=\bigcup_{n=1}^{\infty}Q_n$
Also $\mu=\sum_n \mu_n$.
Now how can i use this technique to prove the second equality which involves the infimum?
Can someone help me with this?
Thank you in advance.
Hint: Show that there exist $0<r_1< r_2 < \dots \to \infty$ such that $\mu(\{x:|x|= r_k\}) = 0.$ Then consider the open sets $B(0,r_1), \{ r_1< |x| < r_2\}, \{ r_2< |x| < r_3\}, \dots.$