Marcinkiewicz integral Let compact set $E \subset (a,b) \subset \mathbb{R}$, and $\lambda >0$. Consider the integral
$M_{F,\lambda}(x)=\int_{a}^{b}\frac{\delta_E^\lambda(y)}{|x-y|^{1+\lambda}}dy$. 
Where $\delta_E(x)=dist(x,E).$
Prove that:
1) If $x \in E^c,$   $M_{F,\lambda}(x)=\infty$
2)$\int_EM_{F,\lambda}(x)dx<\infty$
Any hints? I used fubini's theorem for the 2nd but I have a problem with bounding the integral.Thank you in advance!
 A: Factor dist(y,E) after you applied Fubini's theorem and observe that the function under the first integral is now bounded by 1.
A: Let $F$ be a closed subset of $(a,b)$ and $G=F^c$.  Let $\delta(y) = \rho(y,F)=\inf_{x\in F} \rho(x,y)$ be the distance from $y$ to $F$.  If we plot the graph $\delta(y)$, we see that its slope is either 0, -1, or 1, so it is clearly Lipshitz.  We define the Marcinkiewicz integral to be
\begin{align*}
  M(x) = M_\lambda (x;F) &= \int_G \left|\frac{\delta(y)}{{y-x}}\right|^\lambda \frac{1}{\left|y-x\right|} dy.
\end{align*}
Theorem
$\int_{F}^{} M_\lambda \le 2 |G| \lambda^{-1}$.  Consequently, $M$ is finite a.e. in $F$.
Proof.
Throughout this proof, $x$ refers to a point in $G$ and $y$ refers to a point in $F$.  The idea is to use Tonelli's theorem on the double integral $\int_F M(x)dx = \int_F\int_G \left|\frac{\delta(y)}{{y-x}}\right|^\lambda \frac{1}{\left|y-x\right|} dy dx$.  We have
\begin{align*}
  \int_F M(x) dx &= \int_G \delta^\lambda(y) \left(\int_F  {\left|y-x\right|}^{-1-\lambda} dx\right) dy\\
  &\le \int_G \delta^\lambda(y) \left(\int_{\delta(y)\le\left|y-x\right| }  {\left|y-x\right|}^{-1-\lambda} dx\right) dy\\
  &= \int_G \delta^\lambda(y) \left(\int_{\delta(y)\le t} t^{-1-\lambda} dx\right) dy\\
  &= \int_G \delta^\lambda(y) 2\lambda^{-1} \delta^{-\lambda}(y) dy\\
  &= \int_G 2\lambda^{-1} dy = 2\lambda^{-1} \left|G\right|.
\end{align*}
The inequality is due to $\forall x\in F,\delta(y)\le\left|y-x\right|$ which comes from the defintion $\delta(y) = \inf_{x\in F} \rho(x,y)$.
Theorem
Let $z\in G$.  Then $M(z)=\infty$.
Proof.
$G$ is open, so $z$ is an interior point of $G$.  Find an open interval $I=(z- \epsilon, z+ \epsilon)$ with $G\supset I\ni z$.  On $I$, $\delta(y) \ge  \delta(z)-\epsilon>0$ on $I$, as the clear from the graph of $\delta(y)$ (it has slope either 1 or -1).
We are interested in the integral $\int_G \delta(y) {\left|y-z\right|}^{-1-\lambda} dy \ge \int_I \delta(y) {\left|y-z\right|}^{-1-\lambda} dy.$
\begin{align*}
\int_I \delta(y) {\left|y-z\right|}^{-1-\lambda} dy &\ge (\delta(z)- \epsilon) \int_{(z- \epsilon, z+ \epsilon)} {\left|y-z\right|}^{-1-\lambda} dy \\
&=  2(\delta(z)- \epsilon)    \int_{0}^{\epsilon} t^{-1-\lambda} dy=\infty.
  \end{align*}
