Show that $T$ is a $(p,p)$-opeator. 
Let $\varepsilon>0$, $f\in L^p(\mathbb{R})$ ($1\le p\le \infty$). Opeator $T$ is defined as 
  $$
Tf(x)=\int_{|x-y|>1}\frac{f(y)}{|x-y|^{n+\varepsilon}}\,\mathrm{d}y.
$$
  Show that $T$ is a weak $(1,1)$-opeator and $\color{red}{(p,q)}$-opeator.

Weak $(p,q)$-opeator: Let $1\le p,q\le \infty$, $\mathscr{F}$ be the space of measurable functions on $\mathbb{R}^n$. $T$ is an opeator from $L^p(\mathbb{R}^n)$ to $\mathscr{F}$, and $T$ satisfies:
(i) $\sup_{\alpha>0}\alpha\cdot|\{x\in\mathbb{R}^n: |T(f(x))|>\alpha\}|^{\frac{1}{q}}\le C \lVert f \rVert_p, 1\le q< \infty$;
(ii) $\lVert T(f)\rVert_q\le C\lVert f \rVert_p, q=\infty$.
$(p,q)$-opeator: Let $T$ be an opeator from $L^p(\mathbb{R}^n)$ to $L^q(\mathbb{R}^n)$, $1\le p, q\le \infty$. $T$ is a $(p,q)$-opeator if there exists a constant $C>0$, such that for every $f\in L^p(\mathbb{R}^n)$, $\lVert T(f) \rVert _q\le C\lVert f \rVert_p$.

Since $(p,q)$-opeator is always weak $(p,q)$-opeator, we only need to show that $T$ is a $(p,q)$-opeator, i.e. there exists a constant $C>0$, such that for every $f\in L^p(\mathbb{R})$, we have
$$
\left(\int_{\mathbb{R}}\left|\int_{|x-y|>1}\frac{f(y)}{|x-y|^{n+\varepsilon}}\,\mathrm{d}y\right|^q \mathrm{d}x\right)^{\frac{1}{q}}\le C\left(\int_{\mathbb{R}}|f(x)|^p\,\mathrm{d}x\right)^{\frac{1}{p}}, 1\le q \le \infty.
$$
Could you please help me? Thank you!

Edit: Show that $T$ is a weak $(1,1)$-opeator and $(p,q)$-opeator.
 $\Longrightarrow$ Show that $T$ is a weak $(1,1)$-opeator and $(p,p)$-opeator.
 A: Let $k(x)=\chi_{\{|x|>1\}}|x|^{-n-\varepsilon}$, $x\in \mathbb{R}^n$. We first show that $k(x)$ is in $L^1(\mathbb{R}^n)$.
Consider 
$$
\psi(x)=\sup_{|y|\ge|x|}|k(y)|=
\begin{cases}
 1,&  |x|<1;\\
 k\left( x \right),&  |x|\ge 1.\\
\end{cases}
$$
Then $$
\int_{\mathbb{R}^n}\psi(x)\,\mathrm{d}x=\int_{|x|<1}1\,\mathrm{d}x+\int_{|x|\ge 1}k(x)\,\mathrm{d}x=1+\int_{|x|\ge 1}\frac{\mathrm{d}x}{|x|^{n+\varepsilon}}\stackrel{{?}}{<}\infty \color{red}{\text{ (problem)}}
$$
So $\psi(x)\in L^1(\mathbb{R}^n)$. Also $k(x)\in L^1(\mathbb{R}^n)$ since $0\le k(x)\le \psi(x)$.
Note that $$
Tf(x)=\int_{|x-y|>1}\frac{f(y)}{|x-y|^{n+\varepsilon}}\,\mathrm{d}y=\int_{\mathbb{R}^n}f(y)k(x-y)\,\mathrm{d}y=(f*k)(x),
$$
and $f\in L^p(\mathbb{R}^n)$ $(1\le p \le \infty)$, $k\in L^1(\mathbb{R}^n)$, hence $f*g$ and so $Tf$ is measurable. Then there exists a constant $c_n$ such that
$$
\sup_{r>0}|(f*k_r)(x)|\le c_n \parallel\psi\parallel_1 Mf(x), \forall f\in L^p(\mathbb{R}^n), (1\le p\le \infty),
$$
where $k_r(x)=\frac{1}{r^n}k(\frac{x}{r})$, $M$ is Hardy-Littlewood operator.
Thus, $$
|Tf(x)|=|(f*g_1)(x)|\le c_n \parallel\psi\parallel_1 Mf(x), \forall f\in L^p(\mathbb{R}^n), (1\le p\le \infty).
$$
Since $M$ is weak $(1,1)$-operator and $(p,p)$-operator, so is $T$.
Could you please help me to check and complete this proof? Thanks!
