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A positive Borel measure on the unit disc satisfies $\int |f|d\mu \le N\Vert f{\Vert }_1$ for every H1 space on the unit disk function exactly when it is a Carleson measure, with comparable optimal constants. The normalized squared Cauchy kernel tests necessity. For sufficiency, the boundary shadow of every point with $|f(z)|>t$ lies inside a component arc of $\{Nf>t\}$. The geodesic-cap and Carleson-box comparison bounds the measure of each shadow region by its component length, giving $\mu \{|f|>t\}\le C\Vert \mu {\Vert }_{\mathrm{Carl}}m\{Nf>t\}$. The layer cake representation and H1 non-tangential maximal inequality on the disk give the embedding. The broader kernel and covering mechanism is discussed in Terence Tao's harmonic-analysis notes, Exercise 2 after Lemma 4.1; the analytic endpoint here uses the fractional-power argument just given.

We prove necessity of the geometric condition from the embedding condition. Use normalized angular measure $m$ and the H1 space on the unit disk norm $\Vert f{\Vert }_1={\sup }_{r<1}\int |f(r\xi )|dm(\xi )$. Testing $f=1$ first gives $\mu (\mathbb{D})\le N(\mu )$.

For a short arc $I$ of normalized length $\ell =m(I)$ centered at ${\xi }_0$, put $a=(1-\ell ){\xi }_0$ and use the normalized squared Cauchy kernelIts boundary modulus is the Poisson kernel on the circle, so $\Vert f_a{\Vert }_1=1$. The geodesic-cap and Carleson-box comparison gives $|z-{\xi }_0|\le C\ell$ for $z\in Q(I)$. Therefore $|1-\overline{a}z|\le C^{\prime }\ell$, whereas $1-|a{|}^2=2\ell -{\ell }^2\ge \ell$. Thus $|f_a(z)|\ge c/\ell$ throughout the hyperbolic geodesic cap. The embedding bound givesFor arcs longer than a fixed small threshold, the total-mass estimate gives the same form after changing the absolute constant. HenceThis is precisely the Carleson measure condition, in its geodesic-cap version.