[proofplan]
The bridge between BMO and Carleson measures is the [Littlewood-Paley Decomposition Identity](/theorems/3186):
\begin{align*}
\int_{\mathbb{R}^{n+1}_+}|\nabla u|^2\,t\, d\mathcal{L}^n\,d\mathcal{L}^1 = \tfrac{1}{2}\|f\|_{L^2}^2,
\end{align*}
which we derive inline from [Plancherel](/theorems/247). For the BMO $\Rightarrow$ Carleson direction (proved in full): for a fixed cube $Q$, write $f = (f - f_Q)\mathbb{1}_{2Q} + (f - f_Q)\mathbb{1}_{(2Q)^c} + f_Q$. The constant $f_Q$ contributes nothing to $\nabla u$. The first piece is in $L^2$ with $\|(f - f_Q)\mathbb{1}_{2Q}\|_{L^2}^2 \le C_n\|f\|_{\mathrm{BMO}_2}^2|Q|$ by [Equivalence of BMO $L^p$ Norms](/theorems/3181). The second piece is far from the tent and contributes only via off-diagonal kernel decay. Combining gives $\mu_f(T(Q)) \lesssim \|f\|_{\mathrm{BMO}}^2|Q|$. **The reverse direction (Carleson $\Rightarrow$ BMO) is treated as a sketch:** the standard proof, due to Fefferman--Stein (Acta Math. 129, 1972), uses $H^1$--BMO duality combined with tent-space embedding theorems. We outline the strategy and cite Fefferman--Stein for the full execution; this direction is not given in elementary form here.
[/proofplan]
[step:Set up the harmonic extension and derive the square-function identity from Plancherel]
The Poisson kernel $P_t$ is even, smooth on $\mathbb{R}^n$, satisfies $\int_{\mathbb{R}^n}P_t\, d\mathcal{L}^n = 1$ for every $t > 0$, and the function
\begin{align*}
u : \mathbb{R}^{n+1}_+ &\to \mathbb{R} \\
(x, t) &\mapsto (P_t * f)(x) = \int_{\mathbb{R}^n}P_t(x - y)f(y)\, d\mathcal{L}^n(y)
\end{align*}
is harmonic on $\mathbb{R}^{n+1}_+ = \mathbb{R}^n\times(0, \infty)$, i.e. $\Delta_{(x,t)}u = 0$. The Fourier transform of $P_t$ is $\widehat{P_t}(\xi) = e^{-2\pi t|\xi|}$ (the standard formula for the Poisson semigroup), so for $f \in L^2(\mathbb{R}^n)$, by [Plancherel](/theorems/247),
\begin{align*}
u(x, t) = \int_{\mathbb{R}^n}\hat{f}(\xi)e^{-2\pi t|\xi|}e^{2\pi i\xi\cdot x}\, d\mathcal{L}^n(\xi).
\end{align*}
The full gradient is $\nabla u = (\nabla_x u, \partial_t u)$, with squared modulus
\begin{align*}
|\nabla u|^2 = |\nabla_x u|^2 + |\partial_t u|^2.
\end{align*}
**Computation of $|\nabla_x u|^2$ via Plancherel.** Differentiating under the Fourier integral,
\begin{align*}
\nabla_x u(x,t) = \int_{\mathbb{R}^n}(2\pi i\xi)\hat f(\xi)e^{-2\pi t|\xi|}e^{2\pi i\xi\cdot x}\,d\mathcal{L}^n(\xi),
\end{align*}
so the inverse Fourier transform of $\widehat{\nabla_x u(\cdot, t)}(\xi) = (2\pi i\xi)\hat f(\xi)e^{-2\pi t|\xi|}$. By [Plancherel](/theorems/247),
\begin{align*}
\int_{\mathbb{R}^n}|\nabla_x u(x,t)|^2\,d\mathcal{L}^n(x) = \int_{\mathbb{R}^n}|2\pi\xi|^2|\hat f(\xi)|^2 e^{-4\pi t|\xi|}\,d\mathcal{L}^n(\xi) = (2\pi)^2\int_{\mathbb{R}^n}|\xi|^2|\hat f(\xi)|^2 e^{-4\pi t|\xi|}\,d\mathcal{L}^n(\xi).
\end{align*}
**Computation of $|\partial_t u|^2$ via Plancherel.** Differentiating in $t$,
\begin{align*}
\partial_t u(x,t) = \int_{\mathbb{R}^n}\hat f(\xi)\cdot(-2\pi|\xi|)e^{-2\pi t|\xi|}e^{2\pi i\xi\cdot x}\,d\mathcal{L}^n(\xi),
\end{align*}
so $\widehat{\partial_t u(\cdot, t)}(\xi) = -2\pi|\xi|\hat f(\xi)e^{-2\pi t|\xi|}$. By [Plancherel](/theorems/247),
\begin{align*}
\int_{\mathbb{R}^n}|\partial_t u(x,t)|^2\,d\mathcal{L}^n(x) = (2\pi)^2\int_{\mathbb{R}^n}|\xi|^2|\hat f(\xi)|^2 e^{-4\pi t|\xi|}\,d\mathcal{L}^n(\xi).
\end{align*}
**Sum and integration in $t$.** Both contributions equal $(2\pi)^2\int|\xi|^2|\hat f(\xi)|^2 e^{-4\pi t|\xi|}\,d\mathcal{L}^n(\xi)$, so by addition,
\begin{align*}
\int_{\mathbb{R}^n}|\nabla u(x,t)|^2\,d\mathcal{L}^n(x) = 2(2\pi)^2\int_{\mathbb{R}^n}|\xi|^2|\hat f(\xi)|^2 e^{-4\pi t|\xi|}\,d\mathcal{L}^n(\xi).
\end{align*}
Multiplying by $t$ and integrating $t\in(0,\infty)$, using [Fubini](/theorems/513) (Tonelli for the non-negative integrand),
\begin{align*}
\int_{\mathbb{R}^{n+1}_+}|\nabla u|^2\,t\,d\mathcal{L}^n\,d\mathcal{L}^1 &= 2(2\pi)^2\int_{\mathbb{R}^n}|\xi|^2|\hat f(\xi)|^2\biggl(\int_0^\infty t\,e^{-4\pi t|\xi|}\,d\mathcal{L}^1(t)\biggr)\,d\mathcal{L}^n(\xi).
\end{align*}
The inner $t$-integral is computed by the substitution $s = 4\pi t|\xi|$ (so $t = s/(4\pi|\xi|)$, $dt = ds/(4\pi|\xi|)$):
\begin{align*}
\int_0^\infty t\,e^{-4\pi t|\xi|}\,d\mathcal{L}^1(t) = \frac{1}{(4\pi|\xi|)^2}\int_0^\infty s\,e^{-s}\,d\mathcal{L}^1(s) = \frac{1}{(4\pi|\xi|)^2}\cdot\Gamma(2) = \frac{1}{16\pi^2|\xi|^2}.
\end{align*}
Substituting,
\begin{align*}
\int_{\mathbb{R}^{n+1}_+}|\nabla u|^2\,t\,d\mathcal{L}^n\,d\mathcal{L}^1 = 2(2\pi)^2\cdot\frac{1}{16\pi^2}\int_{\mathbb{R}^n}|\xi|^2|\hat f(\xi)|^2\cdot|\xi|^{-2}\,d\mathcal{L}^n(\xi) = \frac{1}{2}\int_{\mathbb{R}^n}|\hat f(\xi)|^2\,d\mathcal{L}^n(\xi) = \frac{1}{2}\|f\|_{L^2}^2,
\end{align*}
the last by [Plancherel](/theorems/247) again. This is the **Littlewood--Paley square-function identity**:
\begin{align*}
\int_{\mathbb{R}^{n+1}_+}|\nabla u(x, t)|^2 \, t\, d\mathcal{L}^n(x)\,d\mathcal{L}^1(t) = \tfrac{1}{2}\|f\|_{L^2(\mathbb{R}^n)}^2 \tag{LP}
\end{align*}
for every $f\in L^2(\mathbb{R}^n)$ — equivalently, the identity recorded in the [Littlewood-Paley Decomposition Identity](/theorems/3186), with constant $1/2$ in the present normalisation.
[/step]
[step:Direction BMO $\Rightarrow$ Carleson — split $f$ and exploit cancellation]
Assume $f \in \mathrm{BMO}(\mathbb{R}^n)\cap L^2(\mathbb{R}^n)$. Fix a cube $Q \subset \mathbb{R}^n$ with side length $\ell(Q)$ and centre $x_Q$. Let $T(Q) := Q\times(0, \ell(Q)]$ be the **tent** over $Q$, and let $2Q$ denote the cube concentric with $Q$ of side $2\ell(Q)$, so $|2Q| = 2^n|Q|$.
Decompose
\begin{align*}
f = (f - f_Q)\mathbb{1}_{2Q} + (f - f_Q)\mathbb{1}_{(2Q)^c} + f_Q =: g_1 + g_2 + f_Q,
\end{align*}
where the constant $f_Q$ has zero gradient under the harmonic extension (the harmonic extension of a constant is the constant itself). Therefore $\nabla u = \nabla u_1 + \nabla u_2$ with $u_j := P_t * g_j$. By the elementary inequality $|a+b|^2 \le 2(|a|^2 + |b|^2)$,
\begin{align*}
\mu_f(T(Q)) = \int_{T(Q)}|\nabla u|^2 t\, d\mathcal{L}^n\, d\mathcal{L}^1 \le 2\int_{T(Q)}|\nabla u_1|^2 t\, d\mathcal{L}^n\, d\mathcal{L}^1 + 2\int_{T(Q)}|\nabla u_2|^2 t\, d\mathcal{L}^n\, d\mathcal{L}^1.
\end{align*}
**Local piece $u_1$.** Since $g_1 = (f - f_Q)\mathbb{1}_{2Q} \in L^2(\mathbb{R}^n)$ with
\begin{align*}
\|g_1\|_{L^2}^2 = \int_{2Q}|f - f_Q|^2\, d\mathcal{L}^n,
\end{align*}
applying (LP) from step 1 to $g_1$ yields
\begin{align*}
\int_{\mathbb{R}^{n+1}_+}|\nabla u_1|^2 t\, d\mathcal{L}^n\, d\mathcal{L}^1 = \tfrac{1}{2}\|g_1\|_{L^2}^2 = \tfrac{1}{2}\int_{2Q}|f - f_Q|^2\, d\mathcal{L}^n.
\end{align*}
By the [Equivalence of BMO $L^p$ Norms](/theorems/3181) with $p = 2$, $\frac{1}{|2Q|}\int_{2Q}|f - f_{2Q}|^2\, d\mathcal{L}^n \le C_{2,n}^2\|f\|_{\mathrm{BMO}}^2$. The triangle inequality gives $|f - f_Q| \le |f - f_{2Q}| + |f_{2Q} - f_Q|$, with
\begin{align*}
|f_{2Q} - f_Q| = \biggl|\frac{1}{|Q|}\int_Q(f - f_{2Q})\,d\mathcal{L}^n\biggr|\le\frac{1}{|Q|}\int_Q|f - f_{2Q}|\,d\mathcal{L}^n\le\frac{|2Q|}{|Q|}\cdot\frac{1}{|2Q|}\int_{2Q}|f - f_{2Q}|\,d\mathcal{L}^n\le 2^n\|f\|_{\mathrm{BMO}}.
\end{align*}
Hence
\begin{align*}
\int_{2Q}|f - f_Q|^2\, d\mathcal{L}^n &\le 2\int_{2Q}|f - f_{2Q}|^2\, d\mathcal{L}^n + 2|f_{2Q} - f_Q|^2|2Q| \\
&\le 2\,C_{2,n}^2\|f\|_{\mathrm{BMO}}^2|2Q| + 2\cdot 4^n\|f\|_{\mathrm{BMO}}^2|2Q| \\
&= 2(C_{2,n}^2 + 4^n)\cdot 2^n|Q|\,\|f\|_{\mathrm{BMO}}^2.
\end{align*}
Setting $C_n^{(1)} := 2^{n+1}(C_{2,n}^2 + 4^n)$, we obtain $\int_{\mathbb{R}^{n+1}_+}|\nabla u_1|^2 t\, d\mathcal{L}^n\, d\mathcal{L}^1\le\tfrac{1}{2}C_n^{(1)}\|f\|_{\mathrm{BMO}}^2|Q|$. Restricting to $T(Q) \subseteq \mathbb{R}^{n+1}_+$,
\begin{align*}
\int_{T(Q)}|\nabla u_1|^2 t\, d\mathcal{L}^n\, d\mathcal{L}^1\le\tfrac{1}{2}C_n^{(1)}\|f\|_{\mathrm{BMO}}^2|Q|.
\end{align*}
**Far piece $u_2$.** Since $g_2 = (f - f_Q)\mathbb{1}_{(2Q)^c}$ is supported away from the tent, we use the gradient kernel estimate derived from $P_t(z) = c_n t(|z|^2 + t^2)^{-(n+1)/2}$:
\begin{align*}
|\nabla_{(x,t)}P_t(x - y)|\le\frac{C_n}{(|x - y|^2 + t^2)^{(n+1)/2}}\le\frac{C_n}{(|x-y| + t)^{n+1}},
\end{align*}
where $C_n > 0$ is a dimensional constant. (Direct computation: $\partial_{x_i}P_t = -c_n(n+1)t(x_i - y_i)(|z|^2+t^2)^{-(n+3)/2}$ at $z = x-y$, of order $|z|^{-(n+1)}\cdot(t/(|z|+t))$; $\partial_t P_t = c_n(|z|^2+t^2)^{-(n+1)/2} - c_n(n+1)t^2(|z|^2+t^2)^{-(n+3)/2}$, of order $(|z|+t)^{-(n+1)}$. Combining gives the stated bound.)
For $(x, t)\in T(Q)$ and $y\in(2Q)^c$: $x\in Q$ has $|x - x_Q|_\infty\le\ell(Q)/2$ (using the $\ell^\infty$-norm appropriate for cubes), and $y\notin 2Q$ has $|y - x_Q|_\infty\ge\ell(Q)$, so $|x - y|_\infty\ge|y - x_Q|_\infty - |x - x_Q|_\infty\ge\ell(Q)/2$. Converting to the Euclidean norm via the standard inequality $|z|_\infty\le|z|\le\sqrt n\,|z|_\infty$,
\begin{align*}
|x - y|\ge|x - y|_\infty\ge\ell(Q)/2.
\end{align*}
This is a clean dimension-independent bound: we work throughout with the $\ell^\infty$ geometry on cubes (so "$2Q$" denotes the cube of $\ell^\infty$-side $2\ell(Q)$, consistent with the [Calderón--Zygmund Decomposition](/theorems/3154) convention used in the writer's standard cube definitions). Also $|x - y|\asymp|x_Q - y|$ uniformly for $y\in(2Q)^c$ and $x\in Q$: by the same $\ell^\infty$-argument, $|x_Q - y|_\infty\ge\ell(Q)\ge 2|x - x_Q|_\infty$, so $|x_Q - y|/2\ge|x - x_Q|$ via the same inequality chain, and $|x - y|\ge|x_Q - y|/2$. Combining with $t\le\ell(Q)\le|x_Q - y|_\infty\le|x_Q - y|$,
\begin{align*}
|\nabla u_2(x, t)|\le\int_{(2Q)^c}|\nabla_{(x,t)}P_t(x - y)|\,|f(y) - f_Q|\, d\mathcal{L}^n(y)\le C_n\int_{(2Q)^c}\frac{|f(y) - f_Q|}{|x_Q - y|^{n+1}}\, d\mathcal{L}^n(y).
\end{align*}
By a dyadic annular decomposition $(2Q)^c = \bigsqcup_{k\ge 1}(2^{k+1}Q\setminus 2^kQ)$ and the BMO doubling estimate $|f_{2^kQ} - f_Q|\le C_n k\|f\|_{\mathrm{BMO}}$ (induction on $k$ via the one-step estimate above),
\begin{align*}
\int_{(2Q)^c}\frac{|f(y) - f_Q|}{|x_Q - y|^{n+1}}\, d\mathcal{L}^n(y)&\le\sum_{k\ge 1}\frac{1}{(2^k\ell(Q))^{n+1}}\int_{2^{k+1}Q}|f(y) - f_Q|\, d\mathcal{L}^n(y)\\
&\le\sum_{k\ge 1}\frac{1}{(2^k\ell(Q))^{n+1}}\biggl(\int_{2^{k+1}Q}|f - f_{2^{k+1}Q}|\,d\mathcal{L}^n + |f_{2^{k+1}Q} - f_Q|\cdot|2^{k+1}Q|\biggr)\\
&\le\sum_{k\ge 1}\frac{C_n(k+1)\|f\|_{\mathrm{BMO}}\cdot 2^{(k+1)n}\ell(Q)^n}{(2^k\ell(Q))^{n+1}}\\
&= \frac{C_n\|f\|_{\mathrm{BMO}}}{\ell(Q)}\sum_{k\ge 1}(k+1)2^{n - k},
\end{align*}
where the last sum converges geometrically and yields a dimensional constant. Hence
\begin{align*}
|\nabla u_2(x, t)|\le\frac{C_n\|f\|_{\mathrm{BMO}}}{\ell(Q)}\qquad\text{for all }(x, t)\in T(Q).
\end{align*}
Integrating $|\nabla u_2|^2 t$ over $T(Q) = Q\times(0, \ell(Q)]$,
\begin{align*}
\int_{T(Q)}|\nabla u_2|^2 t\, d\mathcal{L}^n\, d\mathcal{L}^1\le\frac{C_n^2\|f\|_{\mathrm{BMO}}^2}{\ell(Q)^2}\cdot|Q|\cdot\int_0^{\ell(Q)}t\, d\mathcal{L}^1(t) = \frac{C_n^2\|f\|_{\mathrm{BMO}}^2}{\ell(Q)^2}\cdot|Q|\cdot\frac{\ell(Q)^2}{2} = \tfrac{1}{2}C_n^2\|f\|_{\mathrm{BMO}}^2|Q|.
\end{align*}
**Combination.** Setting $C_n^{(\Rightarrow)} := C_n^{(1)} + C_n^2$,
\begin{align*}
\mu_f(T(Q))\le 2\cdot\tfrac{1}{2}C_n^{(1)}\|f\|_{\mathrm{BMO}}^2|Q| + 2\cdot\tfrac{1}{2}C_n^2\|f\|_{\mathrm{BMO}}^2|Q| = C_n^{(\Rightarrow)}\|f\|_{\mathrm{BMO}}^2|Q|,
\end{align*}
the Carleson bound. The constant $C_n^{(\Rightarrow)}$ depends only on $n$, and the bound is uniform in $Q$.
[/step]
[step:Direction Carleson $\Rightarrow$ BMO — cite Fefferman--Stein via $H^1$--BMO duality]
Assume $f\in L^2(\mathbb{R}^n)$ and $\mu_f$ is a Carleson measure with norm
\begin{align*}
\|\mu_f\|_{\mathcal{C}} := \sup_Q\frac{\mu_f(T(Q))}{|Q|}.
\end{align*}
We must show $\|f\|_{\mathrm{BMO}}\le C_n^{(\Leftarrow)}\|\mu_f\|_{\mathcal{C}}^{1/2}$ for a dimensional constant $C_n^{(\Leftarrow)}$.
This direction of the equivalence is the harder one. Unlike the BMO $\Rightarrow$ Carleson direction (step 2), which is essentially elementary once the Littlewood--Paley square-function identity (LP) is in hand, the Carleson $\Rightarrow$ BMO direction is **not elementary**. The standard proof is due to **Fefferman and Stein**, *$H^p$ spaces of several variables*, **Acta Mathematica 129** (1972), 137--193, where it appears as part of the broader theorem identifying $\mathrm{BMO}(\mathbb{R}^n)$ as the dual of the real Hardy space $H^1(\mathbb{R}^n)$. We outline the strategy below, give the key estimates in their canonical form, and cite Fefferman--Stein for the technical execution.
**Strategy.** The proof proceeds by **$H^1$--BMO duality** combined with **tent-space embedding theorems**.
(i) *Define a linear functional via the Carleson measure.* Given $\mu_f$ Carleson, define
\begin{align*}
L: H^1(\mathbb{R}^n) &\to \mathbb{C} \\
\varphi &\mapsto \int_{\mathbb{R}^{n+1}_+}(P_t * \varphi)(x)\,t\,d\mu_f(x, t),
\end{align*}
where $P_t * \varphi$ is the Poisson extension of $\varphi$ to the upper half-space. (Heuristically, $L(\varphi) = \langle f, \varphi\rangle$ once we identify $f$ with the boundary trace of its Poisson extension, but at this stage $f$ is fixed and we are constructing a BMO bound from $\mu_f$.)
(ii) *Bound $|L(\varphi)|$ by $\|\mu_f\|_{\mathcal{C}}\|\varphi\|_{H^1}$.* This is a **tent-space duality** estimate: for any atom $\varphi$ supported in a ball $B$, the Poisson extension $P_t * \varphi$ is concentrated on the tent $T(B)$ in a quantitative sense, and pairing with the Carleson measure $\mu_f$ on $T(B)$ produces a bound proportional to $\|\mu_f\|_{\mathcal{C}}|B|^{1/2}\cdot\|\varphi\|_{L^2}\le\|\mu_f\|_{\mathcal{C}}$ for an $H^1$-atom. The full argument requires the **Coifman--Meyer--Stein tent-space theory** (the $T^p_q$-spaces) and the duality $T^1_2(\mathbb{R}^{n+1}_+)^* = T^\infty_2(\mathbb{R}^{n+1}_+)$. See Fefferman--Stein (1972) §11 and §12, and Coifman--Meyer--Stein, *Some new function spaces and their applications to harmonic analysis*, J. Funct. Anal. **62** (1985), 304--335, for the modern formulation.
(iii) *Identify the BMO function via $H^1$--BMO duality.* By [Fefferman's theorem](/theorems/3186) on $H^1$-BMO duality (Fefferman--Stein 1972, Theorem 2), the dual of $H^1(\mathbb{R}^n)$ is $\mathrm{BMO}(\mathbb{R}^n)$, with $\|F\|_{\mathrm{BMO}}\asymp\sup\{|L(\varphi)| : \|\varphi\|_{H^1}\le 1\}$ for $L$ corresponding to $F$. Hence the bounded linear functional $L$ from (ii) corresponds to a function $F\in\mathrm{BMO}(\mathbb{R}^n)$ with $\|F\|_{\mathrm{BMO}}\le C_n\|\mu_f\|_{\mathcal{C}}$.
(iv) *Identify $F$ with $f$.* The key remaining step is to verify that the function $F$ produced by the duality construction coincides with the original $f\in L^2$ — i.e., that the Carleson measure $\mu_F = \mu_f$. This follows by uniqueness of the Poisson extension and the fact that both $f$ and $F$ produce the same Carleson measure $\mu_f$, modulo additive constants. Concretely, the identity $\mu_F(T(Q)) = \int_{T(Q)}|\nabla(P_t * F)|^2\,t\,d\mathcal{L}^n\,d\mathcal{L}^1 = \mu_f(T(Q))$ for every cube $Q$ implies $|\nabla(P_t * F)|^2\,t = |\nabla(P_t * f)|^2\,t$ as measures on $\mathbb{R}^{n+1}_+$, hence $F - f$ has Poisson extension with vanishing gradient, hence $F - f$ is constant. Since BMO is defined modulo constants, $\|F\|_{\mathrm{BMO}} = \|f\|_{\mathrm{BMO}}$, and we obtain
\begin{align*}
\|f\|_{\mathrm{BMO}}\le C_n\|\mu_f\|_{\mathcal{C}}^{1/2}.
\end{align*}
(The square-root appears because $\mu_f$ has dimension of $|f|^2$ — see step 2 — so passing from $\mu_f$ to $f$ in BMO involves taking a square root.)
**This proof is a sketch.** The full execution of (i)--(iv) — particularly the tent-space duality estimate in (ii) — requires substantial machinery from real-variable Hardy space theory beyond the scope of an inline argument. We refer the reader to Fefferman--Stein (1972) for the original proof (where it appears as the harder direction in their characterisation of $\mathrm{BMO} = (H^1)^*$), or to Stein's *Harmonic Analysis* (1993), Chapter IV, §4, for a textbook treatment using tent spaces.
[/step]
[step:Combine the two directions to conclude $\|\mu_f\|_{\mathcal{C}}\asymp \|f\|_{\mathrm{BMO}}^2$]
Combining steps 2 and 3, we have shown the existence of dimensional constants $c_n^{(1)}, c_n^{(2)} > 0$ such that for every $f \in L^2(\mathbb{R}^n)$,
\begin{align*}
\|\mu_f\|_{\mathcal{C}} \le c_n^{(1)}\|f\|_{\mathrm{BMO}}^2 \quad \text{and} \quad \|f\|_{\mathrm{BMO}}^2 \le c_n^{(2)}\|\mu_f\|_{\mathcal{C}}.
\end{align*}
The first inequality is the BMO $\Rightarrow$ Carleson direction with $\|\mu_f\|_{\mathcal{C}} = \sup_Q\mu_f(T(Q))/|Q|$. The second is Carleson $\Rightarrow$ BMO. Together they assert that $\|\mu_f\|_{\mathcal{C}}$ and $\|f\|_{\mathrm{BMO}}^2$ are equivalent quantities, with constants depending only on the dimension $n$. In particular, $f \in \mathrm{BMO}(\mathbb{R}^n)$ if and only if $\mu_f$ is a Carleson measure. This is the conclusion of the theorem.
[/step]
