0$ and that
$$
p\in\biggr(\frac{2-\sqrt{3}}{4}, \, \frac{2+\sqrt{3}}{4}\biggr)
\equiv I_{0} \Leftrightarrow L(p)>0
$$
($I_0$ is about $(0.07,0.93)$).
\proclaim{Theorem 6.6} {\rom{(1)}} $d\mu_p$ has exact dimension $H(p)$.
{\rom{(2)}} Suppose $p\in I_0$. Then for a.e.~$\lambda$ w.r.t.~Lebesgue
measure, the restriction to $[0,1]$ of the rank one perturbation
of $d\mu_p$ has exact dimension $L(p)$.
{\rom{(3)}} If $p\notin\bar I_{0}$, then for a.e.~$\lambda$, the rank
one perturbation of $d\mu_p$ is pure point.
{\rom{(4)} If $p\in(\frac{1}{4}, \frac{3}{4})$, $p\neq\frac{1}{2}$,
then for {\rom{all}} $\lambda$, the restriction to $[0,1]$ of
the rank one perturbation of $d\mu_p$ is purely singular continuous
\rom(so we have an example with singular continuous spectrum for all
$\lambda$\rom).
\endproclaim
\remark{Remarks} 1. (4) says for $p\in (\frac{1}{4}, \frac{3}{4})$,
$G(x)=\infty$ for all $x\in [0,1]$.
2. We'll prove this theorem in Appendix 5.
\endremark
\example{Example 4} {\it{Examples with pure point spectrum}}
\endexample
Our last class of examples will show $\{x\mid G(x)<\infty\}$ can have
any Hausdorff dimension, and also provide examples where $d\mu_\lambda$
has a singular continuous component for all $\lambda\neq 0$ but
sometimes mixed with embedded point spectrum. In this example, $d\mu$
will be a measure fixed once and for all with $\text{supp}(\mu)=
[0,1]$ and $G_{\mu}(x)\equiv\int\frac{d\mu(y)}{(x-y)^2} =\infty$ on
$[0,1]$. Three possibilities to keep in mind are:
\roster
\item"{(1)}" $\chi_{[0,1]}(x)\,dx$ which is absolutely continuous.
\item"{(2)}" $d\mu_{p}$, the measure of Example 3, with $p\in (\frac
{1}{4}, \frac{1}{2})$ where $G(x)=\infty$ by Theorem 6.6(4).
\item"{(3)}" Any of the point measures $d\nu_\alpha$ of Example 2 having
$G(x_{0})=\lim\limits_{\epsilon\downarrow 0}\,\epsilon^{-1}\text{Im}\,
F_{\nu_{\alpha}}(x_{0}+i\epsilon)=\infty$ for all $x_{0}\in [0,1]$.
\endroster
These show there are such $\mu$ with any spectral type.
\proclaim{Theorem 6.7} Let $C$ be an arbitrary closed nowhere dense
set in $[0,1]$. Let $\mu$ be a Borel measure on $[0,1]$ with $G_{\mu}
(x)=\infty$ on $[0,1]$ and $\int d\mu(x)=1$. Let
$$
d\nu(x)=\text{\rom{dist}}(x,C)^{2}\,d\mu(x).
$$
Then, $\text{\rom{supp}}(\nu)=[0,1]$, $G_{\nu}(x)=\infty$ on
$[0,1]\backslash C$ and $G_{\nu}(x)\leq 1$ on $C$.
\endproclaim
\demo{Proof} If $x\notin C$, $\text{dist}(x,C)=\delta >0$ since $C$
is closed. Thus, $G_{\nu}(x)\geq (\frac{\delta}{2})^{2}\int
\limits_{|x-y|\leq\delta/2}\frac{d\mu(y)}{(x-y)^2}\mathbreak =\infty$
since $G_{\mu}(x)=\infty$. On the other hand, if $x\in C$,
$$
G_{\nu}(x)=\int\frac{\text{dist}(y,C)^2}{\text{dist}(x,y)^2}\,d\mu(y)
\leq\int d\mu(y)=1
$$
since $\text{dist}(x,y)\geq\text{dist}(C,y)$. Finally, since $[0,1]
\backslash C$ is dense, $\text{supp}(d\nu)=[0,1]$. \qed
\enddemo
Now let $\tilde\nu$ be $\nu/[\int d\nu]$. Then for every $x\in C$,
$G_{\tilde\nu}(x)\leq\frac{1}{N}$ for $N=\int d\nu$. Consider now
the rank one perturbation $d\tilde{\nu}_{\lambda}$ of $d\tilde{\nu}$.
>From (5.3), we see each pure point has weight at least $\frac{N}
{\lambda^{2}}$ so there are at most $\frac{\lambda^2}{N}$ pure points
(since $d\tilde{\nu}_{\lambda}$ is normalized in (5.3)). Thus,
\proclaim{Proposition 6.8} If $N=\int d\nu(x)$ for the measure $\nu$ of
Theorem {\rom{6.7}}, then $A_{\lambda}\equiv A+\lambda(\varphi, \,\cdot
\,)\varphi$ has at most $\frac{\lambda^2}{N}$ eigenvalues in $[0,1]$. In
particular, if $\lambda^{2}~~0$,
$$
|\{E\mid F(E+i0)>t\}|=|\{E\mid F(E+i0)<-t\}|=t^{-1}\mu(\Bbb R).
$$
\endproclaim
\demo{Proof} Without loss, we can suppose $\mu(\Bbb R)=1$. Let $A_0$
be the operator of multiplication by $x$ on $L^{2}(\Bbb R,d\mu)$ and
$(P\psi)=(1,\psi)1$. Let $d\mu_\lambda$ be the spectral measure for
$A_{0}+\lambda P$. As noted above:
$$
\int d\lambda [d\mu_{\lambda}(E)]=dE
$$
in the sense that for any measurable set $S$,
$$
\int \mu_{\lambda}(S)\,d\lambda =|S|. \tag A.9
$$
On the other hand, by the Aronszajn-Donoghue theory [\svln],
$$
\mu_{\lambda}\text{ is supported on }\{E\mid F(E+i0)=
-\lambda^{-1}\}. \tag A.10
$$
Let $S_{t}=\{E\mid F(E+i0)<-t\}$. Then (A.10) says that
$$
\mu_{\lambda}(S_{t}) =\cases 1, & 0<\lambda~~