- n ^{- \alpha }} {n ^{1- \alpha } } f (x-n) 
\\&
:= \sum_{k=1} ^{\infty }  T_k f (x). \end{aligned}$$ Above, we have passed directly to the distinct scales of the operator. We will subsequently write $ Y_n =  X_n - n ^{- \alpha }$, which are independent mean zero random variables.

The crux of the matter are these two estimates:

\[l:almost\] Almost surely, for all $ 0< \epsilon < 1$, and for all integers $ k$, and $ f, g$ supported on an interval $ I$ of length $ 2 ^k$, we have $$\label{e:almost}
\lvert  \langle   T_k f, g \rangle\rvert 
\lesssim 
\begin{cases}
2 ^{- k \frac {1- \alpha  } 2 + \epsilon  }  \langle f  \rangle _{I,2} \langle g \rangle _{I,2} \lvert  I\rvert 
\\
2 ^{k \alpha } \langle f  \rangle _{I,1} \langle g \rangle _{I,1} \lvert  I\rvert 
\end{cases}.$$ The implied constant is random, but independent of $ k \in \mathbb N $ and the choice of functions $ f,g $.

The second bound follows trivially from $   \lvert  Y_n\rvert /n ^{1- \alpha }  \mathbf 1_{   2 ^{k-1} \leq \lvert  n\rvert < 2 ^{k} }  \lesssim 2 ^{k (\alpha-1)} $. For the first bound, we clearly have $$\lvert  