[proofplan] The proof uses the Calderón-Zygmund decomposition at level $\lambda$ to split $f = g + b$, where the good part $g$ has $L^2$ control and the bad part $b = \sum_j b_j$ is a sum of mean-zero pieces supported on disjoint intervals $Q_j$. We bound the level set of $|Hf|$ by separately controlling $|Hg|$ and $|Hb|$. The good part is handled by the $L^2$ isometry of $H$ combined with Chebyshev's inequality. The bad part is handled by exploiting the cancellation $\int_{Q_j} b_j\, d\mathcal{L}^1 = 0$ together with the Hörmander smoothness condition for the kernel $K(x) = 1/(\pi x)$, which controls the $L^1$ norm of $Hb_j$ outside the doubled interval $2Q_j$. The doubled intervals $\bigcup_j 2Q_j$ have total measure $\lesssim \|f\|_{L^1}/\lambda$, absorbing the unestimated part. [/proofplan]

[step:Apply the Calderón-Zygmund decomposition at level $\lambda$] Fix $f \in L^1(\mathbb{R})$ and $\lambda > 0$. We may assume $f \ge 0$ (otherwise split $f = (\operatorname{Re} f)^+ - (\operatorname{Re} f)^- + i((\operatorname{Im} f)^+ - (\operatorname{Im} f)^-)$, apply the bound to each non-negative summand, and absorb a factor of $4$ into the final constant). Apply the [Calderón-Zygmund Decomposition](/theorems/???) to $f$ at level $\lambda$. The decomposition produces: (a) a countable family of pairwise disjoint open intervals $\{Q_j\}_{j \in J}$ with \begin{align*} \lambda < \frac{1}{|Q_j|}\int_{Q_j} f\, d\mathcal{L}^1 \le 2\lambda \quad \text{for each } j \in J; \end{align*} (b) a decomposition $f = g + b$ where $b: \mathbb{R} \to \mathbb{C}$, $b = \sum_{j \in J} b_j$, with \begin{align*} b_j: \mathbb{R} &\to \mathbb{C} \\ y &\mapsto \mathbb{1}_{Q_j}(y)\left(f(y) - \frac{1}{|Q_j|}\int_{Q_j} f\, d\mathcal{L}^1\right); \end{align*} (c) the good part $g: \mathbb{R} \to \mathbb{C}$ defined by \begin{align*} g(y) = f(y)\mathbb{1}_{\mathbb{R} \setminus \bigcup_j Q_j}(y) + \sum_{j \in J}\left(\frac{1}{|Q_j|}\int_{Q_j} f\, d\mathcal{L}^1\right)\mathbb{1}_{Q_j}(y), \end{align*} which satisfies $\|g\|_{L^\infty} \le 2\lambda$ and $\|g\|_{L^1} \le \|f\|_{L^1}$; (d) the total measure bound $\sum_{j \in J} |Q_j| \le \|f\|_{L^1}/\lambda$. We split the level set: \begin{align*} \{|Hf| > \lambda\} \subseteq \{|Hg| > \lambda/2\} \cup \{|Hb| > \lambda/2\}. \end{align*} By subadditivity of Lebesgue measure, \begin{align*} \mathcal{L}^1(\{|Hf| > \lambda\}) \le \mathcal{L}^1(\{|Hg| > \lambda/2\}) + \mathcal{L}^1(\{|Hb| > \lambda/2\}). \end{align*} [/step]

[step:Bound the good part using $L^2$ isometry and Chebyshev] We control the good part $g$ in $L^2$. Since $0 \le g \le 2\lambda$ (the average and the original $f$ are non-negative on the complement of $\bigcup_j Q_j$ where $f \le \lambda$ by the [Lebesgue Differentiation Theorem](/theorems/???) — see the CZ decomposition for the precise pointwise inequality), we have $|g(y)|^2 \le 2\lambda |g(y)|$, so \begin{align*} \|g\|_{L^2}^2 = \int_{\mathbb{R}} |g|^2\, d\mathcal{L}^1 \le 2\lambda \int_{\mathbb{R}} |g|\, d\mathcal{L}^1 = 2\lambda \|g\|_{L^1} \le 2\lambda \|f\|_{L^1}. \end{align*} By the [$L^2$ Isometry of the Hilbert Transform](/theorems/???), $\|Hg\|_{L^2} = \|g\|_{L^2}$. Chebyshev's inequality gives \begin{align*} \mathcal{L}^1(\{|Hg| > \lambda/2\}) \le \frac{4}{\lambda^2}\|Hg\|_{L^2}^2 = \frac{4}{\lambda^2}\|g\|_{L^2}^2 \le \frac{4}{\lambda^2} \cdot 2\lambda\|f\|_{L^1} = \frac{8\|f\|_{L^1}}{\lambda}. \end{align*} [/step]

[step:Estimate the bad part outside the doubled intervals using cancellation and the Hörmander condition] Let $Q_j^* := 2Q_j$ denote the interval with the same center as $Q_j$ but twice the length, and set $\Omega^* := \bigcup_{j \in J} Q_j^*$. Then \begin{align*} \mathcal{L}^1(\Omega^*) \le \sum_{j \in J} |Q_j^*| = 2\sum_{j \in J} |Q_j| \le \frac{2\|f\|_{L^1}}{\lambda}. \end{align*} Outside $\Omega^*$, we estimate $|Hb|$ in $L^1$. Let $c_j$ denote the center of $Q_j$. Since $\int_{Q_j} b_j\, d\mathcal{L}^1 = 0$, \begin{align*} Hb_j(x) = \frac{1}{\pi}\int_{Q_j} \frac{b_j(y)}{x - y}\, d\mathcal{L}^1(y) = \frac{1}{\pi}\int_{Q_j} \left(\frac{1}{x-y} - \frac{1}{x - c_j}\right) b_j(y)\, d\mathcal{L}^1(y), \end{align*} where we subtracted the constant $\frac{1}{x - c_j}\int_{Q_j} b_j\, d\mathcal{L}^1 = 0$ (the integral against $b_j$ vanishes by the cancellation property of $b_j$). Note that for $x \notin Q_j^* = 2Q_j$, the kernel $1/(x-y)$ is well-defined for $y \in Q_j$, so no principal value is needed. Taking absolute values and integrating over $x \notin Q_j^*$, \begin{align*} \int_{\mathbb{R}\setminus Q_j^*} |Hb_j(x)|\, d\mathcal{L}^1(x) \le \frac{1}{\pi}\int_{Q_j} |b_j(y)| \int_{\mathbb{R}\setminus Q_j^*} \left|\frac{1}{x-y} - \frac{1}{x-c_j}\right| d\mathcal{L}^1(x)\, d\mathcal{L}^1(y), \end{align*} where we used Tonelli's theorem (integrand non-negative) to exchange the integrals. For $y \in Q_j$ and $x \notin 2Q_j$, set $z := x - c_j$; then $|z| \ge |Q_j|$ and $|y - c_j| \le |Q_j|/2 \le |z|/2$. The Hörmander estimate proved for the kernel $K(x) = 1/(\pi x)$ in the chapter gives \begin{align*} \int_{|z| > 2|y-c_j|} \left|\frac{1}{z - (y - c_j)} - \frac{1}{z}\right| d\mathcal{L}^1(z) \le \frac{4}{1} = 4, \end{align*} and since $\{x \notin 2Q_j\} \subseteq \{|x - c_j| > 2|y - c_j|\}$ for $y \in Q_j$, we obtain \begin{align*} \int_{\mathbb{R}\setminus Q_j^*} \left|\frac{1}{x-y} - \frac{1}{x-c_j}\right| d\mathcal{L}^1(x) \le 4. \end{align*} Therefore \begin{align*} \int_{\mathbb{R}\setminus Q_j^*} |Hb_j|\, d\mathcal{L}^1 \le \frac{4}{\pi}\int_{Q_j} |b_j|\, d\mathcal{L}^1 = \frac{4}{\pi}\|b_j\|_{L^1}. \end{align*} Summing over $j$ and using $\|b_j\|_{L^1} \le 2\int_{Q_j}|f|\, d\mathcal{L}^1$ (the average is bounded by $|Q_j|^{-1}\int_{Q_j}|f|\, d\mathcal{L}^1$, so $|b_j| \le |f| + (|Q_j|^{-1}\int_{Q_j}|f|)\mathbb{1}_{Q_j}$ and $\|b_j\|_{L^1} \le 2\int_{Q_j}|f|$), \begin{align*} \sum_{j \in J} \int_{\mathbb{R}\setminus Q_j^*} |Hb_j|\, d\mathcal{L}^1 \le \frac{8}{\pi}\sum_{j \in J} \int_{Q_j} |f|\, d\mathcal{L}^1 \le \frac{8}{\pi}\|f\|_{L^1}. \end{align*} In particular, since $\mathbb{R}\setminus\Omega^* \subseteq \mathbb{R}\setminus Q_j^*$ for each $j$, \begin{align*} \int_{\mathbb{R}\setminus\Omega^*} |Hb|\, d\mathcal{L}^1 \le \sum_{j \in J} \int_{\mathbb{R}\setminus Q_j^*} |Hb_j|\, d\mathcal{L}^1 \le \frac{8}{\pi}\|f\|_{L^1}. \end{align*} By Chebyshev's inequality, \begin{align*} \mathcal{L}^1(\{x \notin \Omega^* : |Hb(x)| > \lambda/2\}) \le \frac{2}{\lambda}\int_{\mathbb{R}\setminus\Omega^*}|Hb|\, d\mathcal{L}^1 \le \frac{16\|f\|_{L^1}}{\pi\lambda}. \end{align*} [/step]

[step:Combine the bounds and absorb the doubled-cube measure] We assemble the level-set estimate. The level set $\{|Hf| > \lambda\}$ is contained in \begin{align*} \{|Hg| > \lambda/2\} \cup \{|Hb| > \lambda/2\} \subseteq \Omega^* \cup \{|Hg| > \lambda/2\} \cup \big(\{|Hb| > \lambda/2\} \cap (\mathbb{R}\setminus\Omega^*)\big). \end{align*} By Step 1 the doubled cubes have total measure $\mathcal{L}^1(\Omega^*) \le 2\|f\|_{L^1}/\lambda$. By Step 2 the good-part level set has measure at most $8\|f\|_{L^1}/\lambda$. By Step 3 the bad-part level set outside $\Omega^*$ has measure at most $16\|f\|_{L^1}/(\pi\lambda)$. Combining, \begin{align*} \mathcal{L}^1(\{|Hf| > \lambda\}) \le \mathcal{L}^1(\Omega^*) + \mathcal{L}^1(\{|Hg| > \lambda/2\}) + \mathcal{L}^1(\{|Hb| > \lambda/2\} \cap (\mathbb{R}\setminus\Omega^*)) \\ \le \frac{2\|f\|_{L^1}}{\lambda} + \frac{8\|f\|_{L^1}}{\lambda} + \frac{16\|f\|_{L^1}}{\pi\lambda} \le \frac{C\|f\|_{L^1}}{\lambda}, \end{align*} with $C = 10 + 16/\pi$. After absorbing the factor of $4$ from the initial reduction to $f \ge 0$ (replacing $f$ by its real and imaginary parts split into positive/negative components), the universal constant becomes $C \le 4(10 + 16/\pi)$, completing the proof. [/step]