two mutually singular measures witch both support is whole $\mathbf{R}$

Question

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.

Answer 1

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 $$\text{supp}(\mu)=\text{supp}(\nu)=\mathbb{R}.$$

Answer 2

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 $\mathbb{R}$.