[proofplan] The strategy is the **good-lambda inequality** of Fefferman–Stein. We compare the level sets of the [Hardy--Littlewood Maximal Inequality](/theorems/2968) maximal function $Mf$ to those of the sharp maximal function $f^\sharp$ via the [Calderón--Zygmund Decomposition](/theorems/3154): on each stopping cube of $\{Mf > \lambda\}$, the constraint $f^\sharp \le \gamma\lambda$ forces the local oscillation to be small, so the overshoot $\{Mf > 2\lambda\} \cap \{f^\sharp \le \gamma\lambda\}$ has measure controlled by $\gamma$ times $\mathcal{L}^n(\{Mf > \lambda\})$. Iterating this distributional inequality in the layer-cake formula for $\|Mf\|_{L^p}^p$ — bootstrapping from the a priori finiteness of $\|Mf\|_{L^p}$ provided by $f \in L^{p_0}$ — gives $\|Mf\|_{L^p} \le C_p\|f^\sharp\|_{L^p}$. The pointwise bound $|f| \le Mf$ a.e. then yields the conclusion. [/proofplan] [step:Recall definitions and the dyadic stopping decomposition of $\{Mf > \lambda\}$] Let $f \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$. The **Hardy--Littlewood maximal function** is the map \begin{align*} Mf : \mathbb{R}^n &\to [0, \infty] \\ x &\mapsto \sup_{Q \ni x}\frac{1}{|Q|}\int_Q |f(y)|\, d\mathcal{L}^n(y), \end{align*} where the supremum runs over all axis-parallel cubes $Q$ containing $x$. The **Fefferman--Stein sharp maximal function** is \begin{align*} f^\sharp : \mathbb{R}^n &\to [0, \infty] \\ x &\mapsto \sup_{Q \ni x}\frac{1}{|Q|}\int_Q |f(y) - f_Q|\, d\mathcal{L}^n(y), \end{align*} where $f_Q := \frac{1}{|Q|}\int_Q f\, d\mathcal{L}^n$ as in the BMO definition. Both maps are lower semicontinuous (as suprema of continuous-in-$x$ averaging functionals over dyadic cubes containing $x$) and pointwise non-negative. Fix $\lambda > 0$. The set $\Omega_\lambda := \{x \in \mathbb{R}^n : Mf(x) > \lambda\}$ is open. The [Calderón--Zygmund Decomposition](/theorems/3154) provides a (possibly empty) countable family $\{Q_j\}$ of pairwise disjoint dyadic cubes — the **stopping cubes** of $\Omega_\lambda$ — such that \begin{align*} \Omega_\lambda &= \bigsqcup_j Q_j, \\ \lambda < \frac{1}{|Q_j|}\int_{Q_j} |f|\, d\mathcal{L}^n &\le 2^n\lambda \qquad \text{for every } j. \end{align*} The lower bound holds because $Q_j \subseteq \Omega_\lambda$ and the dyadic Calderón-Zygmund construction ensures the stopping cube itself has average $> \lambda$. The upper bound holds because the dyadic parent $\widehat Q_j$ of $Q_j$ is **not** contained in $\Omega_\lambda$: there exists $x_*\in\widehat Q_j\setminus\Omega_\lambda$, where $Mf(x_*)\le\lambda$, so the average over $\widehat Q_j$ is at most $\lambda$. Since $|\widehat Q_j| = 2^n|Q_j|$ and $Q_j\subseteq\widehat Q_j$, the average over $Q_j$ is at most $2^n\lambda$. Crucially, since $\widehat Q_j\not\subseteq\Omega_\lambda$, there exists a point $\hat x_j\in\widehat Q_j\setminus\Omega_\lambda$. We will use this point in the case (b) covering argument below. [/step] [step:Establish the good-lambda inequality $\mathcal{L}^n(\{Mf > A\lambda, f^\sharp \le \gamma\lambda\}) \le C_n\gamma\,\mathcal{L}^n(\{Mf > \lambda\})$] Fix $\gamma > 0$ to be chosen later. Set \begin{align*} A := 2\cdot 4^n, \end{align*} a dimensional constant. This is the choice from Stein's standard treatment (Stein, *Harmonic Analysis*, §IV); we fix $A$ to this single value throughout the proof, and all subsequent constants are computed in terms of it. We claim there exists a constant $C_n > 0$ depending only on the dimension such that for every $\lambda > 0$, \begin{align*} \mathcal{L}^n\!\left(\{x : Mf(x) > A\lambda,\, f^\sharp(x) \le \gamma\lambda\}\right) \le C_n\gamma\,\mathcal{L}^n\!\left(\{x : Mf(x) > \lambda\}\right). \tag{GL} \end{align*} Let $\Omega_\lambda$ and $\{Q_j\}$ be as in step 1. Since $A\ge 1$, $\{Mf > A\lambda\}\subseteq\Omega_\lambda = \bigsqcup_j Q_j$, so the LHS equals \begin{align*} \sum_j\mathcal{L}^n(\{x\in Q_j : Mf(x) > A\lambda,\, f^\sharp(x)\le\gamma\lambda\}). \end{align*} Fix $j$ and assume the set is non-empty — otherwise the contribution is zero. Then there is a point $x_0\in Q_j$ with $f^\sharp(x_0)\le\gamma\lambda$. Let $Q_j^* := 3Q_j$ be the cube concentric with $Q_j$ scaled by $3$, so $|Q_j^*| = 3^n|Q_j|$ and $Q_j\subseteq Q_j^*$ (in particular $x_0\in Q_j^*$). Applying the definition of $f^\sharp$ to the cube $Q_j^*$ (which contains $x_0$), \begin{align*} \frac{1}{|Q_j^*|}\int_{Q_j^*}|f - f_{Q_j^*}|\,d\mathcal{L}^n\le f^\sharp(x_0)\le\gamma\lambda. \tag{$*$} \end{align*} We split $f = f_{Q_j^*} + (f - f_{Q_j^*})$. For $x\in Q_j$ and any cube $R\ni x$, two cases arise: **Case (a): $R\subseteq Q_j^*$.** Here $\mathbb{1}_R\le\mathbb{1}_{Q_j^*}$, so \begin{align*} \frac{1}{|R|}\int_R|f|\,d\mathcal{L}^n = \frac{1}{|R|}\int_R|f|\,\mathbb{1}_{Q_j^*}\,d\mathcal{L}^n\le M(\mathbb{1}_{Q_j^*}f)(x). \end{align*} **Case (b): $R\not\subseteq Q_j^*$.** We write out the covering argument. Since $R$ contains $x\in Q_j$ and $R\not\subseteq Q_j^* = 3Q_j$, the side length satisfies $\ell(R) > \ell(Q_j)$ (otherwise $R$ would have side $\le\ell(Q_j)$ and contain $x\in Q_j$, putting it inside $3Q_j$). Now consider the parent cube $\widehat Q_j$ of $Q_j$ (so $\ell(\widehat Q_j) = 2\ell(Q_j)$). By step 1, there is a point $\hat x_j\in\widehat Q_j\setminus\Omega_\lambda$, i.e. $Mf(\hat x_j)\le\lambda$. We claim $R\cup\widehat Q_j$ is contained in a cube $\tilde R$ of side $\le 2\ell(R) + \ell(\widehat Q_j) = 2\ell(R) + 2\ell(Q_j)\le 4\ell(R)$ (using $\ell(R) > \ell(Q_j)$), with $\hat x_j\in\tilde R$. To see this: $R$ has side $\ell(R)$, and $\widehat Q_j$ has side $2\ell(Q_j) < 2\ell(R)$; both meet $Q_j$ (one contains $x\in Q_j$, the other contains $Q_j$), so their centres are within $\ell(R) + \ell(\widehat Q_j)$ of each other; hence the smallest cube containing both has side at most $\ell(R) + 2(\ell(R) + \ell(\widehat Q_j))\le 4\ell(R)$ (a standard geometric estimate for two cubes overlapping a common point). Therefore $|\tilde R|\le 4^n|R|$. Since $\hat x_j\in\tilde R$ and $Mf(\hat x_j)\le\lambda$, \begin{align*} \frac{1}{|\tilde R|}\int_{\tilde R}|f|\,d\mathcal{L}^n\le Mf(\hat x_j)\le\lambda. \end{align*} Multiplying by $|\tilde R|/|R|\le 4^n$ and using $R\subseteq\tilde R$, \begin{align*} \frac{1}{|R|}\int_R|f|\,d\mathcal{L}^n\le\frac{|\tilde R|}{|R|}\cdot\frac{1}{|\tilde R|}\int_{\tilde R}|f|\,d\mathcal{L}^n\le 4^n\lambda. \tag{$**$} \end{align*} So setting $c_n := 4^n$, in case (b) the average over $R$ is at most $c_n\lambda$. **Combining (a) and (b).** For $x\in Q_j$, \begin{align*} Mf(x)\le\max(c_n\lambda,\,M(\mathbb{1}_{Q_j^*}f)(x)). \end{align*} With $A = 2c_n = 2\cdot 4^n$, the inequality $A > c_n$ holds, so on the set $\{Mf(x) > A\lambda\}\cap Q_j$ we have $Mf(x) > A\lambda > c_n\lambda$, forcing $Mf(x)\le M(\mathbb{1}_{Q_j^*}f)(x)$. Thus on $\{Mf > A\lambda\}\cap Q_j$, \begin{align*} M(\mathbb{1}_{Q_j^*}f)(x) > A\lambda. \tag{$\dagger$} \end{align*} **Bound on $|f_{Q_j^*}|$.** We bound the average $\frac{1}{|Q_j^*|}\int_{Q_j^*}|f|\,d\mathcal{L}^n$ directly. Since $\hat x_j\in\widehat Q_j$ and $Q_j^* = 3Q_j$, both sets are contained in the cube $5Q_j$ centred at $x_{Q_j}$ with side $5\ell(Q_j)$ (note $\widehat Q_j\subseteq 3Q_j\subseteq 5Q_j$ and $Q_j^* = 3Q_j\subseteq 5Q_j$). Setting $\tilde Q := 5Q_j$, we have $\hat x_j\in\tilde Q$ and $Mf(\hat x_j)\le\lambda$, so \begin{align*} \frac{1}{|\tilde Q|}\int_{\tilde Q}|f|\,d\mathcal{L}^n\le Mf(\hat x_j)\le\lambda. \end{align*} Since $Q_j^*\subseteq\tilde Q$ and $|\tilde Q|/|Q_j^*| = (5/3)^n$, \begin{align*} |f_{Q_j^*}|\le\frac{1}{|Q_j^*|}\int_{Q_j^*}|f|\,d\mathcal{L}^n\le\frac{|\tilde Q|}{|Q_j^*|}\cdot\frac{1}{|\tilde Q|}\int_{\tilde Q}|f|\,d\mathcal{L}^n\le\biggl(\frac{5}{3}\biggr)^n\lambda =: c_n'\lambda, \end{align*} with $c_n' := (5/3)^n$ a dimensional constant. Note $c_n' < 4^n = c_n$, so $A = 2c_n > c_n + c_n' > c_n'$, and the constants are consistent. **Splitting argument.** Use $\mathbb{1}_{Q_j^*}f = \mathbb{1}_{Q_j^*}f_{Q_j^*} + \mathbb{1}_{Q_j^*}(f - f_{Q_j^*})$. For the constant piece, $M(\mathbb{1}_{Q_j^*}f_{Q_j^*})(x)\le|f_{Q_j^*}|\le c_n'\lambda$ pointwise. Combined with ($\dagger$) on $\{Mf > A\lambda\}\cap Q_j$, \begin{align*} M(\mathbb{1}_{Q_j^*}(f - f_{Q_j^*}))(x)\ge M(\mathbb{1}_{Q_j^*}f)(x) - M(\mathbb{1}_{Q_j^*}f_{Q_j^*})(x) > A\lambda - c_n'\lambda\ge(2c_n - c_n')\lambda\ge c_n\lambda\ge\lambda, \end{align*} where the last steps use $A = 2c_n$, $c_n' < c_n$, and $c_n\ge 1$. By the [Weak-Type Estimate for the Maximal Function](/theorems/2972) applied to $\mathbb{1}_{Q_j^*}(f - f_{Q_j^*})\in L^1(\mathbb{R}^n)$, \begin{align*} \mathcal{L}^n(\{Mf > A\lambda\}\cap Q_j)\le\mathcal{L}^n(\{M(\mathbb{1}_{Q_j^*}(f - f_{Q_j^*})) > \lambda\})\le\frac{C_n^{(W)}}{\lambda}\int_{Q_j^*}|f - f_{Q_j^*}|\,d\mathcal{L}^n, \end{align*} where $C_n^{(W)} > 0$ is the weak-type $(1,1)$ constant. By ($*$), $\int_{Q_j^*}|f - f_{Q_j^*}|\,d\mathcal{L}^n\le\gamma\lambda|Q_j^*| = 3^n\gamma\lambda|Q_j|$, hence \begin{align*} \mathcal{L}^n(\{Mf > A\lambda\}\cap Q_j)\le\frac{C_n^{(W)}}{\lambda}\cdot 3^n\gamma\lambda|Q_j| = 3^n C_n^{(W)}\gamma|Q_j|. \end{align*} Setting $C_n := 3^n C_n^{(W)}$ and summing over $j$, \begin{align*} \mathcal{L}^n(\{Mf > A\lambda,\,f^\sharp\le\gamma\lambda\})\le\sum_j 3^n C_n^{(W)}\gamma|Q_j| = C_n\gamma|\Omega_\lambda|, \end{align*} which is (GL). [/step] [step:Bootstrap to a finite a priori $L^p$ norm of $Mf$ from the integrability hypothesis] Fix $1 < p < \infty$. We will need a quantitative starting point: the $L^p$ norm of $Mf$ must be **finite** before we can integrate it against the good-lambda inequality. We supply this by truncation and use the strong-type $(p, p)$ bound for $M$ from the [Hardy--Littlewood Maximal Inequality](/theorems/2968): \begin{align*} \|Mg\|_{L^p(\mathbb{R}^n)} \le A_p\|g\|_{L^p(\mathbb{R}^n)} \qquad \text{for all } g \in L^p(\mathbb{R}^n), \ 1 < p \le \infty, \end{align*} with $A_p$ depending only on $p$ and $n$. For $N > 0$ and $R > 0$, set $f_{N,R} := f\cdot\mathbb{1}_{B(0, R)\cap\{|f|\le N\}}$. Then $f_{N,R}\in L^\infty\cap L^1$ with bounded support, hence $f_{N,R}\in L^p(\mathbb{R}^n)$ for every $p\in[1,\infty]$. By the Hardy-Littlewood bound, $\|Mf_{N,R}\|_{L^p}\le A_p\|f_{N,R}\|_{L^p} < \infty$. We will prove the sharp inequality \begin{align*} \|f_{N,R}\|_{L^p}\le C_p\|f_{N,R}^\sharp\|_{L^p} \tag{T} \end{align*} uniformly in $N, R$, then pass to the limit $N, R\to\infty$ at the end (step 5). In what follows, we write $f$ for the truncated $f_{N,R}$ to lighten notation; the resulting estimate (T) is uniform in the truncation parameters. [/step] [step:Iterate the good-lambda inequality in the layer-cake formula for $\|Mf\|_{L^p}^p$] By the [Layer-Cake Formula](/theorems/2956), \begin{align*} \|Mf\|_{L^p}^p = \int_{\mathbb{R}^n}(Mf)^p\, d\mathcal{L}^n = \int_0^\infty p\lambda^{p-1}\mathcal{L}^n(\{Mf > \lambda\})\, d\mathcal{L}^1(\lambda). \end{align*} Substituting $\lambda = A\mu$ — where $A$ is the dimensional constant from (GL) and the substitution is an increasing diffeomorphism of $(0, \infty)$ — \begin{align*} \|Mf\|_{L^p}^p = A^p\int_0^\infty p\mu^{p-1}\mathcal{L}^n(\{Mf > A\mu\})\, d\mathcal{L}^1(\mu). \end{align*} Splitting the level set $\{Mf > A\mu\}$ over the disjoint events $\{f^\sharp > \gamma\mu\}$ and $\{f^\sharp \le \gamma\mu\}$ and applying (GL) to the second, \begin{align*} \mathcal{L}^n(\{Mf > A\mu\}) \le \mathcal{L}^n(\{f^\sharp > \gamma\mu\}) + C_n\gamma\,\mathcal{L}^n(\{Mf > \mu\}). \end{align*} Multiplying by $p\mu^{p-1}$ and integrating over $\mu\in(0,\infty)$, \begin{align*} \int_0^\infty p\mu^{p-1}\mathcal{L}^n(\{Mf > A\mu\})\, d\mathcal{L}^1(\mu) &\le \int_0^\infty p\mu^{p-1}\mathcal{L}^n(\{f^\sharp > \gamma\mu\})\, d\mathcal{L}^1(\mu) \\ &\quad + C_n\gamma \int_0^\infty p\mu^{p-1}\mathcal{L}^n(\{Mf > \mu\})\, d\mathcal{L}^1(\mu). \end{align*} Substituting $t = \gamma\mu$ in the first integral on the right, \begin{align*} \int_0^\infty p\mu^{p-1}\mathcal{L}^n(\{f^\sharp > \gamma\mu\})\, d\mathcal{L}^1(\mu) = \gamma^{-p}\int_0^\infty pt^{p-1}\mathcal{L}^n(\{f^\sharp > t\})\, d\mathcal{L}^1(t) = \gamma^{-p}\|f^\sharp\|_{L^p}^p. \end{align*} The second integral on the right is exactly $\|Mf\|_{L^p}^p$ by layer-cake. Therefore \begin{align*} \|Mf\|_{L^p}^p\le A^p\gamma^{-p}\|f^\sharp\|_{L^p}^p + A^p C_n\gamma\,\|Mf\|_{L^p}^p. \end{align*} [/step] [step:Absorb the $\|Mf\|_{L^p}$ term and conclude $\|f\|_{L^p}\le C_p\|f^\sharp\|_{L^p}$] Choose $\gamma > 0$ small enough that $A^p C_n\gamma\le\tfrac{1}{2}$, i.e. $\gamma := (2A^pC_n)^{-1}$, where $A$ and $C_n$ are the dimensional constants from (GL). Then the inequality from step 4 reads \begin{align*} \|Mf\|_{L^p}^p\le A^p\gamma^{-p}\|f^\sharp\|_{L^p}^p + \tfrac{1}{2}\|Mf\|_{L^p}^p, \end{align*} and using the bootstrap of step 3 — that $\|Mf\|_{L^p}^p < \infty$ for the truncated $f = f_{N,R}$ — we may subtract $\tfrac{1}{2}\|Mf\|_{L^p}^p$ from both sides: \begin{align*} \tfrac{1}{2}\|Mf\|_{L^p}^p\le A^p\gamma^{-p}\|f^\sharp\|_{L^p}^p, \end{align*} i.e. \begin{align*} \|Mf\|_{L^p}\le 2^{1/p}\,A\gamma^{-1}\|f^\sharp\|_{L^p} = 2^{1/p}A\cdot 2A^pC_n\,\|f^\sharp\|_{L^p} = 2^{1+1/p}A^{p+1}C_n\,\|f^\sharp\|_{L^p} =: \widetilde{C}_p\|f^\sharp\|_{L^p}, \end{align*} with $\widetilde C_p := 2^{1+1/p}A^{p+1}C_n$ depending only on $p$ and $n$. By the [Lebesgue Differentiation Theorem](/theorems/74), $|f(x)|\le Mf(x)$ for $\mathcal{L}^n$-a.e. $x\in\mathbb{R}^n$ (the average of $|f|$ over shrinking cubes converges to $|f(x)|$ a.e.), hence \begin{align*} \|f\|_{L^p}\le\|Mf\|_{L^p}\le\widetilde{C}_p\|f^\sharp\|_{L^p}. \end{align*} This establishes (T) for the truncation $f = f_{N,R}$. **Removing the truncation.** Let $f\in L^{p_0}(\mathbb{R}^n)$ for some $p_0\in(1,\infty)$ be the original function. We must show $\|f\|_{L^p}\le C_p\|f^\sharp\|_{L^p}$. We treat the cases $p\le p_0$ and $p > p_0$ independently. **Pointwise sharp-function comparison.** For any truncation $g$ of $f$ (a measurable function with $|g|\le|f|$), we have the pointwise bound \begin{align*} g^\sharp(x)\le f^\sharp(x) + 2M(|f - g|)(x)\qquad\text{for every } x\in\mathbb{R}^n. \tag{$\ddagger$} \end{align*} Indeed, for any cube $Q\ni x$, the elementary identity $g - g_Q = (f - f_Q) - (f - g) + (f - g)_Q$ together with $|(f - g)_Q|\le\frac{1}{|Q|}\int_Q|f - g|\,d\mathcal{L}^n\le M(|f - g|)(x)$ gives \begin{align*} \frac{1}{|Q|}\int_Q|g - g_Q|\,d\mathcal{L}^n\le\frac{1}{|Q|}\int_Q|f - f_Q|\,d\mathcal{L}^n + 2M(|f - g|)(x), \end{align*} and ($\ddagger$) follows by taking the supremum over cubes $Q\ni x$. **Case $p\le p_0$ via weak-type $(p_0, p_0)$.** Set $\tilde f_N: \mathbb{R}^n\to\mathbb{C}$, $x\mapsto f(x)\mathbb{1}_{\{|f|\le N\}}(x)$, the level-set truncation (no spatial cutoff). Then $|\tilde f_N|\le|f|\in L^{p_0}$, $|\tilde f_N|\nearrow|f|$ pointwise a.e. as $N\to\infty$, and $\tilde f_N\in L^{p_0}\cap L^\infty$, hence $\tilde f_N\in L^p$ for every $p\in[p_0, \infty]$ (by interpolation, since $\tilde f_N$ is bounded). For $p\le p_0$ we further note $\tilde f_N$ may not be in $L^p$ globally, so we use the **doubly truncated** version $f_{N,R}: \mathbb{R}^n\to\mathbb{C}$, $x\mapsto f(x)\,\mathbb{1}_{B(0,R)\cap\{|f|\le N\}}(x)$, which lies in $L^p\cap L^\infty$ for every $p$ (bounded support, bounded values). For $p\le p_0$: by the strong-type $(p_0, p_0)$ bound from the [Hardy--Littlewood Maximal Inequality](/theorems/2968), \begin{align*} \|M(|f - f_{N,R}|)\|_{L^{p_0}(\mathbb{R}^n)}\le A_{p_0}\|f - f_{N,R}\|_{L^{p_0}(\mathbb{R}^n)}. \end{align*} Since $|f - f_{N,R}| = |f|\mathbb{1}_{E_{N,R}}$ where $E_{N,R} := (B(0,R)\cap\{|f|\le N\})^c$ has $\mathbb{1}_{E_{N,R}}\to 0$ a.e. as $N,R\to\infty$, and $|f - f_{N,R}|\le|f|\in L^{p_0}$, [Dominated Convergence](/theorems/4) gives $\|f - f_{N,R}\|_{L^{p_0}}\to 0$. By the **weak-type $(p_0, p_0)$** estimate (Chebyshev applied to the strong-type bound), $M(|f - f_{N,R}|)\to 0$ in $L^{p_0}$, hence — passing to a subsequence indexed by $N = R\to\infty$ — $M(|f - f_{N,R}|)\to 0$ pointwise a.e. By ($\ddagger$), \begin{align*} \limsup_{N,R\to\infty}f_{N,R}^\sharp(x)\le f^\sharp(x)\qquad\text{for a.e. } x\in\mathbb{R}^n. \end{align*} Apply (T) to $f_{N,R}\in L^p\cap L^\infty$: \begin{align*} \|f_{N,R}\|_{L^p(\mathbb{R}^n)}\le\widetilde C_p\,\|f_{N,R}^\sharp\|_{L^p(\mathbb{R}^n)}. \end{align*} By [Monotone Convergence](/theorems/509) applied to $|f_{N,R}|^p\nearrow|f|^p$ on the LHS, and Fatou's lemma applied to the non-negative sequence $(f_{N,R}^\sharp)^p$ on the RHS (using the pointwise limsup bound just derived), \begin{align*} \|f\|_{L^p}^p = \lim_{N,R\to\infty}\|f_{N,R}\|_{L^p}^p\le\widetilde C_p^p\liminf_{N,R\to\infty}\|f_{N,R}^\sharp\|_{L^p}^p\le\widetilde C_p^p\int_{\mathbb{R}^n}\liminf_{N,R}(f_{N,R}^\sharp)^p\,d\mathcal{L}^n\le\widetilde C_p^p\,\|f^\sharp\|_{L^p}^p. \end{align*} This gives $\|f\|_{L^p}\le\widetilde C_p\|f^\sharp\|_{L^p}$ for $p\le p_0$. **Case $p > p_0$ via strong-type $(p, p)$ on truncations.** For this case, use the level-set truncation $\tilde f_N$ (no spatial cutoff). Since $\tilde f_N$ is bounded ($|\tilde f_N|\le N$) and $|\tilde f_N|\le|f|\in L^{p_0}$, we have \begin{align*} \|\tilde f_N\|_{L^p}^p = \int_{\{|f|\le N\}}|f|^p\,d\mathcal{L}^n\le N^{p - p_0}\int_{\{|f|\le N\}}|f|^{p_0}\,d\mathcal{L}^n\le N^{p - p_0}\|f\|_{L^{p_0}}^{p_0} < \infty, \end{align*} where the inequality $|f|^p\le N^{p - p_0}|f|^{p_0}$ on $\{|f|\le N\}$ follows from $|f|^{p - p_0}\le N^{p - p_0}$ (using $p > p_0$). Hence $\tilde f_N\in L^p\cap L^{p_0}$, and the bootstrap of step 3 gives $\|M\tilde f_N\|_{L^p} < \infty$. Therefore (T) — derived under the hypothesis $f\in L^{p_0}$, but actually proved for any function whose maximal $L^p$-norm is finite — applies to $\tilde f_N$: \begin{align*} \|\tilde f_N\|_{L^p}\le\widetilde C_p\,\|\tilde f_N^\sharp\|_{L^p}. \end{align*} By ($\ddagger$) applied to $g := \tilde f_N$, \begin{align*} \tilde f_N^\sharp(x)\le f^\sharp(x) + 2M(|f - \tilde f_N|)(x). \end{align*} Now $|f - \tilde f_N| = |f|\mathbb{1}_{\{|f| > N\}}$ has $L^{p_0}$-norm tending to zero by DCT (since $|f - \tilde f_N|\le|f|\in L^{p_0}$ and $\mathbb{1}_{\{|f| > N\}}\to 0$ a.e.), so by passing to a subsequence $M(|f - \tilde f_N|)\to 0$ pointwise a.e. (via $L^{p_0}$-continuity of $M$). Hence $\limsup_N\tilde f_N^\sharp\le f^\sharp$ a.e. Apply [Monotone Convergence](/theorems/509) to $|\tilde f_N|^p\nearrow|f|^p$ on the LHS, and Fatou's lemma to $(\tilde f_N^\sharp)^p$ on the RHS: \begin{align*} \|f\|_{L^p}^p = \lim_{N\to\infty}\|\tilde f_N\|_{L^p}^p\le\widetilde C_p^p\liminf_{N\to\infty}\|\tilde f_N^\sharp\|_{L^p}^p\le\widetilde C_p^p\int_{\mathbb{R}^n}\liminf_N(\tilde f_N^\sharp)^p\,d\mathcal{L}^n\le\widetilde C_p^p\,\|f^\sharp\|_{L^p}^p. \end{align*} The two outer bounds are independent of any assumption on $\|f\|_{L^p}$: if $\|f^\sharp\|_{L^p} = \infty$, the inequality is vacuous; if $\|f^\sharp\|_{L^p} < \infty$, the chain forces $\|f\|_{L^p}\le\widetilde C_p\|f^\sharp\|_{L^p} < \infty$. **Conclusion.** In both cases, $\|f\|_{L^p}\le\widetilde C_p\|f^\sharp\|_{L^p} =: C_p\|f^\sharp\|_{L^p}$, with $C_p := \widetilde C_p$ depending only on $p$ and $n$. This completes the proof. [/step]