How to prove a function is a regular Borel measure on $\mathbb{R}$

Question

We have a function $w: \mathbb{Q} \rightarrow [0, \infty]$ which obeys the following property: $$ \sum_{r \in \mathbb{Q} \cap (-b, +b)} w(r) < \infty, \ \ \ \ \ \ 0 < b < \infty $$

for $E \subseteq \mathbb{R}$, define the function $\mu(E) = 0$, if $E = \emptyset$ and $$\mu(E) = \sum_{r \in \mathbb{Q} \cap E} w(r), \ \ \ \ \ \ E \neq \emptyset.$$

How do I prove that $\mu$ is a regular Borel measure on $\mathbb{R}$ ?

I tried to relate the measure defined here with the Lebesgue-Stiletjes measure on $\mathbb{R}$, then from that conclude somehow that the function defined above is regular.

P.S: a regular Borel measure satisfies the following:

if $A \subseteq \mathbb{R}$ is measurable and given $\epsilon > 0$, then there is an open set $G \subseteq \mathbb{R}$, such that:

1. $A \subseteq G$.
2. $\mu\left( G \setminus A \right) < \epsilon$.

This is a homework question, I am not asking for solutions, just for hints.

Thank you,

Regards.

Answer 1

Note that $\mu(E)=\sum_{r \in \Bbb{Q}}w(r)\delta_q(E)$ is defined on the powerset of $\Bbb{R}$.

Here $\delta_q$ denotes the Dirac measure.

It is true that if $\mu_n$ is a sequence of measures on a sigma algebra then $\sum_n\mu_n$ is also a measure.(A good exercise for you).From this, by taking an enumeration of the rationals,you can deduce that $\mu$ is a measure,

Since it is defined on the powerset then it is defined on every Borel set.

Now from hypothesis,the Borel measure $\mu$ is locally finite on $\Bbb{R}$,thus regular.

A reference for that can be found at: http://fourier.math.uoc.gr/~papadim/

In the measure theory lecture notes(go to course notes),on page 64,theorem 5.7

Theorem 5.7 is a general theorem for regularity of Borel measures on a topological space.

It would be quite instructive to read its proof ,becuase uses techniques that you will encounter often in measure theory.