We have 2 functions,
$$\alpha(x)=\left\{ \begin{array} {cl} \lfloor x \rfloor + \log(1+x),& x\geq 0 \\ \lfloor x \rfloor, & x<0 \end{array} \right.$$
$$\beta(x)=\lfloor x \rfloor$$
and $\mu_\alpha,\mu_\beta$ their Borel-Stieltjes measures.
- First I have to prove that for every Borel $A$ of $\mathbb{R}$ we have $\mu_\beta=\mathrm{card}(A\cap \mathbb{Z})$
I did it by separing $A$ in intervals of the type $(a_i,b_i]$ and so $$\mu_\beta(A)=\sum (\lfloor b_i \rfloor-\lfloor a_i \rfloor)=\sum \mathrm{card}((a_i,b_i]\cap\mathbb{Z})=\mathrm{card}(A\cap\mathbb{Z})$$
Do you think is that right?
- Then I have to find a function $f:\mathbb{R}\longrightarrow [0,\infty)$ that satisfies $$\mu_\alpha-\mu_\beta=\int_A f dm$$ where m is Lebesgue measure
I found that
$$(\mu_\alpha-\mu_\beta)((a,b])=\left\{ \begin{array} {cl} 0,& b,a< 0 \\ \log(b+1), & a<0,b\geq 0 \\ \log(b+1)-\log(a+1), & a,b\geq 0 \end{array} \right.$$ but I don't know exactly how to use it.
- Finally I have to calculate $\mu_\alpha([1,3]\setminus \mathbb{Q})$ and $\mu_\alpha(C) $ where $C$ is the Cantor set.
I'm wondering if maybe exercise 2 can be used... because this time I cannot used intervals, I think.
Thank you for any help
You proved it for finite union of half open intervals. Now you have to extend this to the Borel sets. This can be done by a monotone class argument.
You have the data $g(a,b):=\int_{(a,b)}f\mathrm dm$ for $a\lt b$, and you want to deduce from this the value of $f$. Since $g(a,b)=0$ for $a\lt b\lt 0$, we have $f(x)=0$ for $x\lt 0$. To determine the value for $x\gt 0$, use $h(b)-h(a)=\int_a^b h'(t)\mathrm dt$ where $h(t)= \log(1+t)$.
Yes, you can use question 2.: if you know $\mu_\beta(B)$ for some Borel set $B$, then you can deduce $\mu_\alpha(B)$. The former is easier to compute.