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I have some problem with an exercise(for homework):

Find two mutually singular measures $u$ and $v$ (Borel finite on $\mathbf{R}$) with $$\mathrm{supp}(u)=\mathrm{supp}(v)=\mathbf{R}.$$

I tried to solve exercises in this way: $$ u(A)= 1 \ \text{ if }\ |A \cap Q|\geq 1,\ \text{ otherwise }\ 0. $$ $$ v(A) =1 \ \text{ if }\ |A \cap I| \geq 1, \ \text{ otherwise }\ 0. $$ It is clear that supp(u) = supp(v) = R. But is $u \perp v?? $. Thank you in advance.

Eric Stucky
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user207950
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  • And what problem would that be? :) – Shaun Jan 14 '15 at 22:16
  • You know :) :) in solving problem :) :) – user207950 Jan 14 '15 at 22:18
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2 Answers2

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Try finding a set $E$ with closure $\mathbb{R}$, for which $E^c$ also has closure $\mathbb{R}$. Then define one measure on $E$ and the other on $E^c$. This way they are mutually singular and they will both have support $\mathbb{R}$.

Titus
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  • Thank's. I thought on problem in that way(but I couldn't find measure) :) – user207950 Jan 14 '15 at 22:21
  • Maybe Q and I= R/Q. I have idea in this way: u(A) = if |A intersect Q|>=1 than 1 otherwise 0. v(A) = if |A intersect I|>=1 than 1 otherwise 0. Do you think that is right conclusion?? – user207950 Jan 14 '15 at 22:24
  • That's close to what I was thinking. I'm not sure what your definition of 'Borel finite' is, but if the measure needs to be finite ($\mu(\mathbb{R}) < \infty$) then you need to make the measure on $I$ decay at infinity. Similar statements for the measure on $\mathbb{Q}$. – Titus Jan 14 '15 at 22:32
  • 'Borel finite'<=> u(E) < \infty for E \in Borel algebra. I don't understand how to make measure decay at infinity. If I make u(A) as previous, measure u(R) =1. Isn't it?? Or?? – user207950 Jan 14 '15 at 22:38
  • Just start with a measure $\mu$ on all reals which is Borel finite (e.g. the gaussian). When you set $\mu$ to zero on the rational numbers you preserve Borel finiteness. – Titus Jan 14 '15 at 23:18
  • To "set a measure to zero on the rational numbers" when the measure is gaussian refers to a concept I cannot fathom. Please explain. (The same remark applies to the answer itself, for example I do not know what "a measure on $\mathbb Q^c$" refers to. But the OP must know, since they accepted this answer immediately?) – Did Jan 15 '15 at 11:04
  • Given your history on the site @Did I get the feeling that you take issue with my answer, rather than fail to understand what I'm getting at. – Titus Jan 15 '15 at 18:45
  • Setting a measure $\mu$ to zero on a Borel set $A$ would refer to having a provided measure $\mu$, which is used to generate a measure $\mu'$ by the following two rules: $\mu'(B)$ should preserve the measure $\mu(B)$ for $B \subset A^c$ and set $\mu'(B) = 0$ for $B \subset A$ (assuming, yes, that $B$ is measurable). – Titus Jan 15 '15 at 18:47
  • For example, the Lebesgue measure is zero on $\mathbb Q\cap[0,1]$? – Did Jan 15 '15 at 18:48
  • If you are taking $\mu$ to be the Lebesgue measure on $\mathbb{R}$ and $\mu'$ to be the Lebesgue measure `set to zero' outside $[0,1]$, then yes, that's an example. However, the measure on the rationals doesn't change when you restrict to a smaller interval. – Titus Jan 15 '15 at 18:58
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Let $r_n:\mathbb{N}\to\mathbb{Q}$ be an enumeration of the rationals, and $s_n:\mathbb{N}\to\pi+\mathbb{Q}$ be an enumeration of the numbers of the form $\pi+{}$(a rational). Let $\delta_a$ be the Dirac measure supported at $x=a$, i.e.

$$\delta_0(A)=\begin{cases}0&\text{ if }a\notin A\\1&\text{ if }a\in A\end{cases}$$

Define $$\begin{align}\mu&:=\sum_{n=1}^{\infty}2^{-n}\delta_{r_n}\\\nu&:=\sum_{n=1}^{\infty}2^{-n}\delta_{s_n}\end{align}$$

To check they are mutually singular consider the sets $A:=\mathbb{Q}\cup((-\infty,0)\setminus(\pi+\mathbb{Q}))$ and $B:=(\pi+Q)\cup([0,+\infty)\setminus\mathbb{Q})$. We have that $A\cup B=\mathbb{R}$, while $\mu(B)=\nu(A)=0$.

Since $\mathbb{Q}$ and $\pi+\mathbb{Q}$ are dense in $\mathbb{R}$ we see that for every point every one of its neighborhoods will have positive measure in both $\mu$ and $\nu$. Therefore $$\text{supp}(\mu)=\text{supp}(\nu)=\mathbb{R}.$$

Pp..
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