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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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    @Leandro Yours is an amazingly simple solution. Please post it as an answer so that we can upvote it. :-)2011-12-16

3 Answers 3

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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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    @tb Indeed. It's very nice. :-)2011-12-16
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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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    @tb, Aw, beat me to it. +1 :)2011-12-16
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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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    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).2011-12-16