Given a finite measure $\mu$ on the real line $\mathbb R$, one definition of its Hilbert transform is $(H\mu)(y) =\frac{1}{\pi}(PV)\int \frac{d\mu(x)}{x-y}$ which is known to exist almost everywhere on $\mathbb R$. Another way is to define the Borel transform ${\mathcal H}(z) = \frac{1}{\pi}\int\frac{d\mu(x)}{x-z}$ for $z\in \mathbb C\setminus \mathbb R$, and then take the limit of $\Re({\mathcal H}\mu)(x+iy)$ as $y\downarrow 0$, such limit existing almost everywhere. In the case that $\mu$ is absolutely continuous it is stated $explicitly$ in the literature that these agree (almost everywhere), and it is $implicit$ in the use of the term `Hilbert transform' in the literature that the two definitions agree (almost everywhere) for general finite $\mu$. Does anyone know where one can find explicit proof(s)? (Proofs for the $L^p$ case are easy to find.)

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    $\begingroup$ Split $\mu$ into a smooth density measure $\mu_1$, a small absolutely continuous measure $\mu_2$ and a singular measure $\mu_3$. Since the definitions agree on $\mu_1$, it is enough to check that the difference is small on $\mu_2$ and $\mu_3$ outside a small measure set. But it is dominated by the (restricted to $r\le r_0$) Hardy-Littlewood maximal function, which is small outside a set of small measure (due to small total mass for $\mu_2$ and due to being supported on a zero measure set for $\mu_3$). In short: just take the classical $L^1$ proof and modify it a tiny bit. $\endgroup$
    – fedja
    Mar 25, 2011 at 12:45
  • $\begingroup$ @Rick: Could you explain why $H(\mu)$ is well defined for arbitrary finite measure $\mu$? $\endgroup$
    – Syang Chen
    Mar 27, 2011 at 9:04
  • $\begingroup$ @Xianghong: this is precisely Rick's question, and fedja's comment sketches why this works. $\endgroup$
    – Yemon Choi
    Mar 27, 2011 at 19:09
  • $\begingroup$ Hi Rick. I've taken the liberty of adding a "reference-request" tag to your question, since I got the impression that you were looking as much for a place where these facts are written down, as for a sketch of why they are true. $\endgroup$
    – Yemon Choi
    Mar 30, 2011 at 23:16

2 Answers 2


The Hilbert transform of $\mu$ is the inverse Fourier transform of the product $$ -i\hat\mu(\xi){\pi \text{sign}\xi}, $$ using the definition $\hat u(\xi)=\int e^{-2i\pi x\cdot \xi} u(x) dx$ so that the inverse Fourier transform is $u(x)=\int e^{2i\pi x\cdot \xi} \hat u(\xi) d\xi$. This product makes sense since $\hat\mu\in L^\infty$.


Showing that the two definitions agree almost everywhere is easy! Using the truncated transform $$ \mathcal{H}\_\epsilon\mu(x)=\frac1\pi\int_{\lvert y-x\rvert > \epsilon}\frac{d\mu(y)}{x-y} $$ then, by definition, $\mathcal{H}\mu(x)=\lim_{\epsilon\to0}\mathcal{H}\_\epsilon\mu(x)$ for all $x$ at which the limit exists. Convolve the identity $$ \Re\left(\frac{1}{x+ih}\right)=\frac{x}{x^2+h^2}=\int_0^11_{\left\lbrace\lvert x\rvert > h\sqrt{t/(1-t)}\right\rbrace}\frac1x\\,dt $$ with $\frac1\pi d\mu$ to obtain, $$ \Re\left(\mathcal{H}\mu\right)(x+ih)=\int_0^1\mathcal{H}_{h\sqrt{t/(1-t)}}\mu(x)\\,dt. $$ The integrand on the right hand side tends to $\mathcal{H}\mu(x)$ as $h\to0$, whenever this is defined, so bounded convergence gives $\Re(\mathcal{H}\mu)(x+ih)\to\mathcal{H}\mu(x)$.

The question also asks how to show that $\mathcal{H}\mu$ is defined almost everywhere. I don't have a reference for this, but the standard proof can be modified without too much difficulty. The maximal operator $\mathcal{H}^*\mu(x)\equiv\sup_{\epsilon > 0}\lvert\mathcal{H}_\epsilon\mu(x)\rvert$ is weak (1,1) continuous, $$ \left\lvert\left\lbrace x\colon\mathcal{H}^*\mu(x) > \lambda\right\rbrace\right\rvert\le C\lVert\mu\rVert/\lambda, $$ for all $\lambda > 0$ and a fixed constant $C$ (I'm using $\lvert\cdot\rvert$ to denote the Lebesgue measure). I'm working from the notes Interpolation, Maximal Operators, and the Hilbert Transform, by Michael Wong (Theorem 8.7). These notes look at the case where $\mu$ is absolutely continuous, but the proof carries across to the general case with no big changes.

By weak continuity, if there exists measures $\mu_n$ with $\lVert\mu-\mu_n\rVert\to0$ such that $\mathcal{H}\mu_n$ all exist almost everywhere, then $\mathcal{H}\mu$ exists almost everywhere. If $\mu$ is absolutely continuous with differentiable density, then $\mathcal{H}\mu$ exists everywhere in the standard way. As the differentiable functions are dense in $L^1$, this extends to all absolutely continuous measures. Only the case for singular measures $\mu$ remain. So, there exists a measurable $S\subseteq\mathbb{R}$ with zero Lebesgue measure with $\mu(S^c)=0$. Then we can choose compact sets $K_n\subseteq S$ with $\mu(S\setminus K_n)\to0$. Note that the measures $\mu_n\equiv 1_{K_n}\cdot\mu$ are supported on the compact sets $K_n$ and $\lVert\mu-\mu_n\rVert\to0$. It follows that $\mathcal{H}\mu_n(x)$ is defined everywhere outside of $K_n$ (in fact, $\mathcal{H}\_\epsilon\mu_n(x)$ is a constant function of $\epsilon$ for $\epsilon$ small enough that $B_\epsilon(x)\cap K_n=\emptyset$). So, $\mathcal{H}\mu_n$ is defined almost everywhere and, hence, so is $\mathcal{H}\mu$.


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