Borel measure on $[0,1]$ s.t. $\mathbb{Q} \cap [0,1]$ has measure $\frac12$ and $\mu([0,1])=1$ I need to define a Borel measure on $[0,1]$ s.t. the set of rational numbers in $[0,1]$ has measure 1/2 and $\mu([0,1]) = 1$.
I know that the interval is "mostly" made up of irrationals and that there is a rational in between every pair of irrationals and vice-a-versa. So this reminds me of a fat Cantor set but am having a bit of trouble linking that thought to a defined Borel measure....
Any help?
Thanks!
 A: How about enumerating $\mathbb{Q} \cap [0,1]$ as $\{q_1, q_2, q_3, \ldots\}$, putting $q_0 = \frac{\sqrt{2}}{2}$ and considering $\mu = \sum_{n \geq 0} 2^{-(n+1)} \delta_{q_n}$ where $\delta_{x}$ is the point measure $\delta_x(A) = 1$ if $x \in A$ and $\delta_x(A) = 0$ (also called Dirac measure — thanks Dylan) otherwise?
Another solution would be to take $\mu = \frac{1}{2} \delta_0 + \frac{1}{2} \lambda$ where $\lambda$ is the Lebesgue measure on $[0,1]$ and a third solution — perhaps the simplest one — would be a convex combination $\frac{1}{2} \delta_x + \frac{1}{2}\delta_y$ with $x$ rational and $y$ irrational, as was pointed out by Leandro in his answer.
A: Following the suggestion of Srivatsan and t.b. I am posting a previous comment here:
An example of Borel measure you are looking for can be given by
$$
\mu = \frac{1}{2}\delta_{\frac{1}{3}}+\frac{1}{2}\delta_{\frac{\pi}{4}}
$$
As t.b. pointed out this is a Borel measure which is also inner and out regular. For these details see this post :
A question about regularity of Borel measures
A: Fix an enumeration of $\mathbb Q \cap [0, 1]$, say, $(q_n)_{n \in \mathbb N}$, and define a measure $\nu$ supported on the rationals such that $\nu(\{q_n\}) = \frac{6}{\pi^2 n^2}$. Now consider a suitable convex combination of $\nu$ and the standard Lebesgue measure on $[0,1]$.
