# Borel measure on $[0,1]$ s.t. $\mathbb{Q} \subset [0,1]$ 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!

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If you just need a Borel Measure you can consider the following measure: $\mu=\frac{1}{2}\delta_{\frac{1}{3}}+\frac{1}{2}\delta_{\frac{\pi}{3}}$, where $\delta_x$ the delta measure concentrated in $x$. I guess that the problem it is more interesting if you want a Borel measure that is both inner regular and outer regular. –  Leandro Dec 16 '11 at 3:08
@Leandro: your measure is both inner regular and outer regular. In fact, every Borel probability measure on a complete separable metric space is inner and outer regular and tight. See also here –  t.b. Dec 16 '11 at 3:13
@t.b. I don't knew that fact. Thanks for the comment. –  Leandro Dec 16 '11 at 3:16
@Leandro Yours is an amazingly simple solution. Please post it as an answer so that we can upvote it. :-) –  Srivatsan Dec 16 '11 at 3:17

## 3 Answers

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

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Thanks all! This solution is the clearest to me - I wish I knew more about the inner and outer regularity (thanks also for the links!) –  nate Dec 16 '11 at 3:42
Just to make sure: checking regularity of that measure is straightforward. You don't need the arguments in the other thread for that, but it is good to know the general statement (which isn't quite as widely known as it deserves to be). –  t.b. Dec 16 '11 at 3:56

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.

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Also known as Dirac measures. –  Dylan Moreland Dec 16 '11 at 3:07
@tb, Aw, beat me to it. +1 :) –  Srivatsan Dec 16 '11 at 3:09

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 $[0,1]$.

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Sorry about that. Leandro's solution in the comments is still simpler :) -- well if he took $\pi/4$ instead of $\pi/3$. –  t.b. Dec 16 '11 at 3:14
@tb Indeed. It's very nice. :-) –  Srivatsan Dec 16 '11 at 3:19