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A positive Borel measure near a domain boundary is a Carleson measure when its mass in boundary neighborhoods is bounded by a constant times their boundary measure. On the unit disc, this becomes the displayed bound for every Carleson box. The equivalent hyperbolic geodesic cap version follows from the geodesic-cap and Carleson-box comparison. The Carleson embedding theorem on the disk explains why this geometric growth condition controls integration of analytic Hardy spaces.

This Hardy space consists of holomorphic functions with $\Vert f{\Vert }_1={\sup }_{r<1}{\int }_{\partial \mathbb{D}}|f(r\xi )|dm(\xi )<\infty$, where $m$ is normalized angular measure. The H1 non-tangential maximal inequality on the disk controls their approach to the boundary, and the Carleson embedding theorem on the disk characterizes measures against which all these functions remain integrable.

For $a\in \mathbb{D}$, this holomorphic function has H1 space on the unit disk norm one: its boundary modulus is the unit-mass Poisson kernel on the circle. The radial integrals are bounded by that boundary integral by subharmonic comparison and tend to it by continuity. Choosing $a$ at depth comparable to an arc's length makes $|f_a|$ bounded below by the reciprocal length throughout its hyperbolic geodesic cap. It tests necessity in the Carleson embedding theorem on the disk.

We prove sufficiency by a non-tangential maximal function and the layer cake representation. Write $K=C(\mu )$. First $\mu (\mathbb{D})\le CK$: three hyperbolic geodesic caps corresponding to arcs of normalized length $2/3$, with equally spaced centers, cover the unit disc. This handles interior points and large arcs, not only a thin boundary strip.

Fix aperture two and set $Nf(\xi )={\sup }_{z\in {\Gamma }_2(\xi )}|f(z)|$. The crucial H1 non-tangential maximal inequality on the disk isHere is its endpoint proof. For $0<R<1$ let $f_R(z)=f(Rz)$, which is holomorphic across the closed unit disc. Since modulus powers of holomorphic functions are subharmonic, Poisson comparison givesThe estimate that a Poisson maximal function is dominated by the Hardy-Littlewood maximal function yields $(N{f}_R{)}^{1/2}\le CM{g}_R$. The strong $L^2$ Hardy-Littlewood maximal inequality now givesFor each interior point $f_R(z)\to f(z)$, hence $Nf\le {\mathrm{liminf}}_{R\uparrow 1}N{f}_R$. The Fatou lemma proves the maximal inequality. This uses a genuine $L^2$ maximal bound; the Hardy-Littlewood maximal function is not strongly bounded on $L^1$.

For $t>0$, the set $O_t=\{\xi :Nf(\xi )>t\}$ is open, since each interior point witnessing this strict inequality has an open boundary shadow. Write its disjoint arc components as $I_j$. If $|f(z)|>t$ and $\delta =1-|z|$, the arc centered at $z/|z|$ with angular half-width $\delta$ lies in $O_t$, because $|z-\xi |<2\delta$ there. Thus it lies in one component $I_j$, and $z$ has angular projection in $I_j$ with $\delta \le \pi m(I_j)$. The geodesic-cap and Carleson-box comparison puts all such points in $Q(\kappa I_j)$ for a fixed enlargement factor $\kappa$ whenever $I_j$ is small. For large $I_j$, use $\mu (\mathbb{D})\le CK\le C^{\prime }Km(I_j)$. If $z=0$ or $O_t$ is the entire circle, that same total-mass estimate applies. Summing, without needing the enlarged caps to be disjoint, givesThe layer cake representation and the proved H1 non-tangential maximal inequality on the disk finish the argument:Thus the two conditions are equivalent, with their least admissible constants comparable: $C(\mu )\le C_{1}N(\mu )$ and $N(\mu )\le C_{2}C(\mu )$. This is the Carleson embedding theorem on the disk, proved at the analytic $H^1$ endpoint.