Complete convergence of the Hilbert transform

Abstract: Suppose that ${a_j}\in \ell^1$, and suppose that for any sequence $(t_n)$ of integers there exits a constant $C_1>0$ such that $$\sharp\left{k\in\mathbb{Z}:\sup_{n\geq 1}\left|\sum_{i\in \mathcal{B}n-t_n} !!!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{a{k+i}}{i}\right|>\lambda\right}\ \leq C_1\sharp\left{k\in\mathbb{Z}:\sup_{n\geq 1}\left|\sum_{i\in \mathcal{B}n} !!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{a{k+i}}{i}\right|>\lambda\right},$$ for all $\lambda >0$, where $\mathcal{B}n={-n, -(n-1), -(n-2),\dots , n-2, n-1, n}$. Then there is a constant $C_2>0$ which does not depend on the sequence ${a_j}$ such that $$\sum{n=1}^\infty\sharp\left{k\in\mathbb{Z}:\left|\sum_{i=-n}^{n} !!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{a_{k+i}}{i}\right|>\lambda\right}\leq\frac{C_2}{\lambda}\sum_{i=-\infty}^{\infty}|a_i|$$ for all $\lambda>0$. Let $(X,\mathscr{B},\mu )$ be a measure space, $\tau :X\to X$ an invertible measure-preserving transformation, and suppose that $f\in L^1(X)$ such that for any sequence $(t_n)$ of integers there exists a constant $C_1>0$ such that $$\mu\left{ x: \sup_{n\geq 1}\left|\sum_{i\in \mathcal{B}n-t_n}!!!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{f(\tau^ix)}{i}\right| >\lambda \right}\leq C_1\mu\left{x: \sup{n\geq 1}\left|\sum_{i\in \mathcal{B}n}!!\raise{1.9ex}\hbox{$\scriptsize\prime$}\; \frac{f(\tau^i x)}{i}\right|>\lambda \right} $$ for all $\lambda >0$, where $\mathcal{B}_n={-n, -(n-1), -(n-2),\dots , n-2, n-1, n}$. Then there exists a constant $C_2>0$ which does not depend on $f$ such that $$\sum{n=1}^\infty\mu\left{x:\left|\sum_{i=-n}^{n}!!\raise{1.9ex}\hbox{$\scriptsize\prime$} \;\frac{f(\tau^ix)}{i}\right|>\lambda\right}\leq\frac{C_2}{\lambda}|f|_1$$ for all $\lambda >0$.