Why is the Calderón–Zygmund lemma the generalization of the Rising Sun lemma?
Below is a commonly encountered lemma about continuous functions, which you can find in books like Stein’s Real Analysis or on related Wikipedia pages. Let’s call it Rising Sun Lemma 1:
Rising Sun Lemma 1: Let $G:\mathbb R\rightarrow \mathbb{R}$ be continuous and $E=\lbrace x:\exists y>x, G(y)>G(x)\rbrace$. Then $E$ is the union of at most countable disjoint open intervals ${(a_k,b_k)}_k$ such that if $(a_k,b_k)$ is a finite interval, then $G(a_k)=G(b_k)$.
If we further assume $G(-\infty)=\infty$, $G(\infty)=-\infty$, then each $(a_k,b_k)$ must be finite. We can prove this by contradiction (and I leave this as an exercise). We now can introduce the second version of Rising Sun Lemma:
Rising Sun Lemma 2: Let $f\in L^1(\mathbb{R})\cap L^+(\mathbb{R})$ and $\lambda>0$. There exists at most countable disjoint intervals $\lbrace I_k\rbrace_k$ such that
- $\forall k$, $\frac{1}{|I_k|}\int_{I_k}f(x)dx=\lambda$
- $f(x)\leq \lambda$ a.e. on $(\bigcup_kI_k)^c$.
This version looks quite similar to the Calderón-Zygmund decomposition. In fact, it’s an even finer result because in the Calderón-Zygmund decomposition, you can only obtain
\[\lambda< \frac{1}{|I_k|}\int_{I_k}f(x)dx\leq 2\lambda\]So how can we prove Rising Sun Lemma 2? Of course it uses Rising Sun Lemma 1. Define the function
\[G(x)=\int_{-\infty}^xf(x)dx-\lambda x\]It’s easy to verify that $G$ is continuous and satisfies $G(-\infty)=\infty$, $G(\infty)=-\infty$ (by the integrability of $f$). Therefore, using Rising Sun Lemma 1, you obtain countably many finite intervals $\lbrace (a_k,b_k) \rbrace_k$ such that $G(a_k)=G(b_k)$, from which we easily deduce the first statement in Rising Sun Lemma 2. By using the Lebesgue Differentiation Theorem, you can further know that the derivative $G’$ exists a.e. and equal to $f(x)-\lambda$. Since $G$ is decreasing on $(\bigcup_k(a_k,b_k))^c$ (this can be understood by looking at the diagram of the Rising Sun Lemma), we conclude that $G’(x)\leq 0$ and $f(x)\leq \lambda$ on $(\bigcup_k(a_k,b_k))^c$.
