[proofplan]
The strategy is to combine [Cotlar's identity](/theorems/???), which dominates the maximal Hilbert transform pointwise by the Hardy-Littlewood maximal function applied to $Hf$ and to $f$, with the $L^p$ boundedness of the [Hardy-Littlewood maximal function](/theorems/???) and of the [Hilbert transform](/theorems/3169). Almost-everywhere convergence then follows by the standard density argument: the convergence $H_\varepsilon f \to Hf$ holds pointwise on the dense subspace $\mathcal{S}(\mathbb{R})$, and the $L^p$ bound on $H_*$ promotes this to a.e. convergence on all of $L^p(\mathbb{R})$.
[/proofplan]
[step:Recall Cotlar's pointwise domination of $H_* f$ by maximal averages]
Define the truncated and maximal Hilbert transforms by
\begin{align*}
H_\varepsilon &: L^p(\mathbb{R}) \to L^p(\mathbb{R}), & H_\varepsilon f(x) &= \frac{1}{\pi}\int_{|x-y|>\varepsilon} \frac{f(y)}{x-y}\, d\mathcal{L}^1(y), \\
H_* &: L^p(\mathbb{R}) \to [0,\infty], & H_* f(x) &= \sup_{\varepsilon > 0} |H_\varepsilon f(x)|,
\end{align*}
and let $M: L^1_{\mathrm{loc}}(\mathbb{R}) \to [0,\infty]$ be the [Hardy-Littlewood maximal operator](/theorems/???),
\begin{align*}
Mf(x) = \sup_{r > 0} \frac{1}{2r}\int_{x-r}^{x+r} |f(y)|\, d\mathcal{L}^1(y).
\end{align*}
[Cotlar's identity](/theorems/???) gives an absolute constant $C_0 > 0$ such that
\begin{align*}
H_* f(x) \le M(Hf)(x) + C_0\, Mf(x) \quad \text{for $\mathcal{L}^1$-a.e. } x \in \mathbb{R}, \quad f \in L^p(\mathbb{R}), \ 1 < p < \infty.
\end{align*}
[/step]
[step:Take $L^p$ norms and apply the boundedness of $M$ and $H$]
Let $1 < p < \infty$ and $f \in L^p(\mathbb{R})$. Taking $L^p$ norms in the Cotlar inequality and applying Minkowski's inequality:
\begin{align*}
\|H_* f\|_{L^p} \le \|M(Hf)\|_{L^p} + C_0\, \|Mf\|_{L^p}.
\end{align*}
The [Hardy-Littlewood maximal theorem](/theorems/???) gives a constant $A_p > 0$ (depending only on $p$) such that $\|Mg\|_{L^p} \le A_p\, \|g\|_{L^p}$ for all $g \in L^p(\mathbb{R})$; this requires $1 < p \le \infty$, which is satisfied. Applied to $g = Hf$ and to $g = f$:
\begin{align*}
\|M(Hf)\|_{L^p} \le A_p\, \|Hf\|_{L^p}, \qquad \|Mf\|_{L^p} \le A_p\, \|f\|_{L^p}.
\end{align*}
The [$L^p$ boundedness of the Hilbert transform](/theorems/3169) gives a constant $B_p > 0$ such that $\|Hf\|_{L^p} \le B_p\, \|f\|_{L^p}$ for $1 < p < \infty$. Combining,
\begin{align*}
\|H_* f\|_{L^p} \le A_p B_p\, \|f\|_{L^p} + C_0 A_p\, \|f\|_{L^p} = C_p\, \|f\|_{L^p}, \quad C_p := A_p(B_p + C_0).
\end{align*}
[/step]
[step:Reduce a.e. convergence to a control on $\limsup_{\varepsilon \to 0}|H_\varepsilon f - Hf|$]
Fix $f \in L^p(\mathbb{R})$. The differences $H_\varepsilon f - Hf$ are bounded pointwise by $|H_\varepsilon f| + |Hf| \le H_* f + |Hf|$, so
\begin{align*}
\limsup_{\varepsilon \to 0^+} |H_\varepsilon f(x) - Hf(x)| \le 2\, H_* f(x) \quad \text{for $\mathcal{L}^1$-a.e. } x \in \mathbb{R}.
\end{align*}
Set $\Lambda f(x) := \limsup_{\varepsilon \to 0^+} |H_\varepsilon f(x) - Hf(x)|$. The map $f \mapsto \Lambda f$ is sublinear on $L^p(\mathbb{R})$: for $f, g \in L^p$,
\begin{align*}
|H_\varepsilon(f+g) - H(f+g)| \le |H_\varepsilon f - Hf| + |H_\varepsilon g - Hg|,
\end{align*}
so $\Lambda(f+g) \le \Lambda f + \Lambda g$. Therefore $\Lambda$ is sublinear and dominated by $2 H_*$, hence by Step 2,
\begin{align*}
\|\Lambda f\|_{L^p} \le 2\, \|H_* f\|_{L^p} \le 2 C_p\, \|f\|_{L^p}.
\end{align*}
[/step]
[step:Verify a.e. convergence on the dense subspace $\mathcal{S}(\mathbb{R})$]
For $g \in \mathcal{S}(\mathbb{R})$ and $x \in \mathbb{R}$ fixed, write
\begin{align*}
H_\varepsilon g(x) - Hg(x) &= -\frac{1}{\pi}\int_{|x-y| < \varepsilon} \frac{g(y) - g(x)}{x - y}\, d\mathcal{L}^1(y),
\end{align*}
where the term $\int_{|x-y|<\varepsilon} g(x)/(x-y)\, d\mathcal{L}^1(y) = 0$ by oddness of $1/(x-y)$ on the symmetric set $\{|x-y|<\varepsilon\}$. By the mean value theorem, $|g(y) - g(x)| \le \|g'\|_\infty\, |y - x|$, so the integrand is bounded by $\|g'\|_\infty$, and
\begin{align*}
|H_\varepsilon g(x) - Hg(x)| \le \frac{1}{\pi}\int_{|x-y|<\varepsilon} \|g'\|_\infty\, d\mathcal{L}^1(y) = \frac{2\varepsilon}{\pi}\, \|g'\|_\infty \to 0
\end{align*}
as $\varepsilon \to 0^+$. Hence $H_\varepsilon g(x) \to Hg(x)$ pointwise (in fact uniformly in $x$), and so $\Lambda g \equiv 0$ on $\mathbb{R}$ for every $g \in \mathcal{S}(\mathbb{R})$.
[/step]
[step:Promote $\Lambda f = 0$ a.e. from $\mathcal{S}(\mathbb{R})$ to all of $L^p(\mathbb{R})$]
Fix $f \in L^p(\mathbb{R})$ and $\lambda > 0$. Since $\mathcal{S}(\mathbb{R})$ is dense in $L^p(\mathbb{R})$, choose $g \in \mathcal{S}(\mathbb{R})$ with $\|f - g\|_{L^p} < \delta$ (the value of $\delta > 0$ to be chosen). By sublinearity of $\Lambda$ (Step 3) and the vanishing $\Lambda g \equiv 0$ (Step 4),
\begin{align*}
\Lambda f(x) \le \Lambda(f - g)(x) + \Lambda g(x) = \Lambda(f - g)(x) \quad \text{a.e.}
\end{align*}
By Chebyshev's inequality and the $L^p$ bound on $\Lambda$:
\begin{align*}
\mathcal{L}^1\bigl(\{x : \Lambda f(x) > \lambda\}\bigr) \le \mathcal{L}^1\bigl(\{x : \Lambda(f-g)(x) > \lambda\}\bigr) \le \frac{\|\Lambda(f-g)\|_{L^p}^p}{\lambda^p} \le \frac{(2 C_p)^p\, \delta^p}{\lambda^p}.
\end{align*}
Letting $\delta \to 0$ gives $\mathcal{L}^1(\{\Lambda f > \lambda\}) = 0$ for every $\lambda > 0$, hence $\Lambda f = 0$ a.e. Therefore $H_\varepsilon f(x) \to Hf(x)$ for $\mathcal{L}^1$-a.e. $x \in \mathbb{R}$.
[/step]
[step:Combine the maximal bound and a.e. convergence to conclude]
Step 2 gives the inequality $\|H_* f\|_{L^p} \le C_p\, \|f\|_{L^p}$ with $C_p = A_p(B_p + C_0)$, and Step 5 gives $H_\varepsilon f(x) \to Hf(x)$ for a.e. $x \in \mathbb{R}$. Both conclusions hold for every $f \in L^p(\mathbb{R})$ and every $1 < p < \infty$, completing the proof.
[/step]
