The Calderón–Zygmund theorem

Step 1 (Calderón–Zygmund decomposition). Fix $f \in L^1$ and $\lambda>0$. Using a stopping-time argument on dyadic cubes (equivalently, the Hardy–Littlewood maximal function $Mf$), split $f = g + b$ where the 'good' part satisfies $\|g\|_\infty \le C\lambda$ and $\|g\|_1 \le \|f\|_1$, and the 'bad' part $b = \sum_j b_j$ is a sum of pieces supported on pairwise disjoint dyadic cubes $Q_j$ with mean zero, $\int_{Q_j} b_j = 0$, and total measure $\sum_j |Q_j| \le \frac{C}{\lambda}\|f\|_1$ (this is exactly the weak-(1,1) bound for $M$ applied to the stopping cubes).

Step 2 (the good part). Since $T$ is bounded on $L^2$ and $\|g\|_2^2 \le \|g\|_\infty \|g\|_1 \le C\lambda\|f\|_1$, Chebyshev's inequality gives $|\{|Tg| > \lambda/2\}| \le \frac{4}{\lambda^2}\|Tg\|_2^2 \le \frac{C}{\lambda}\|f\|_1$: the good part alone already satisfies the weak-(1,1) bound.

Step 3 (the bad part, off the doubled cubes). Let $Q_j^*$ be the cube with the same center as $Q_j$ and twice the side length; the union $\bigcup_j Q_j^*$ has measure at most $\frac{C}{\lambda}\|f\|_1$ as well. Away from $Q_j^*$, the mean-zero condition on $b_j$ lets one subtract a constant from the kernel: $Tb_j(x) = \int_{Q_j} \big(K(x-y)-K(x-c_j)\big) b_j(y)\,dy$ for the center $c_j$ of $Q_j$, and the Hörmander smoothness bound $|K(x)| \le \dfrac{C}{|x|^n}, \qquad |\nabla K(x)| \le \dfrac{C}{|x|^{n+1}}$ shows this difference is small enough that $\int_{(Q_j^*)^c} |Tb_j(x)|\,dx \le C\|b_j\|_1$.

Step 4 (combine). Summing Step 3 over $j$ and applying Chebyshev off the doubled cubes gives $\big|\{x \notin \bigcup_j Q_j^* : |Tb(x)|>\lambda/2\}\big| \le \frac{C}{\lambda}\sum_j \|b_j\|_1 \le \frac{C}{\lambda}\|f\|_1$. Adding the measure of $\bigcup_j Q_j^*$ from Step 1, the bad part contributes $O(\|f\|_1/\lambda)$ to the level set as well, and combining with Step 2 for the good part gives the full estimate $\big|\{x : |Tf(x)| > \lambda\}\big| \le \dfrac{C}{\lambda}\|f\|_{L^1}$.