Uniqueness of measures related to the Stieltjes transforms

Question

Suppose that $\mu$ is a Radon measure on $\mathbb{R}$, and let $I(\mu)$ be the Stieltjes transform of $\mu$, i.e. $I(\mu)(\lambda)=\int_{\mathbb{R}}\frac{1}{x-\lambda}\,\mathrm{d}\mu(x)$ for $\lambda$ non-real. If $I(\mu)(\lambda)$ happens to be equal to zero, would it imply that $\mu=0$? If yes, why? I can't find any theorem stating about it, so I ask for your help. Maybe it has something with injectivity to do with?

Answer 1

As I mentioned in an earlier comment, The imaginary part of the Stieltjes transform $I_\mu$ (in your notation) is $\pi P*\mu$, where $P$ is the Poisson kernel.

A short but concise presentation if the Stieltjes integral in Barry Simon's Comprehensive Course in Analysis, Volume 3 (Harmonic Analysis) section 2.5. Amongst all the different cool things one can find there is the following Theorem:

Theorem: Let $\mu$ be a finite measure on $\mathbb{R}$ and $F_\mu(z)=\int\frac{\mu(dx)}{x-z}$ be its Stieltjes transform. Suppose $$ \mu(dx) = f\cdot\lambda(dx) + \mu_s(dx)$$ be its Lebesgue decomposition (absolute and singular decomposition with respect the LEbeshue measure $\lambda$. Then

a. $\frac{1}{\pi}\int \operatorname{Im}\,F_\mu(x+ i\varepsilon)g(x)\,dx \xrightarrow{\varepsilon\rightarrow0+}\int g(x)\mu(dx)$ for any $g\in \mathcal{C}_{00}(\mathbb{R})$ (functions of compact support).

b. For $\lambda$-a.a. $x\in \mathbb{R}$, $$\frac{1}{\pi}\lim_{\varepsilon\rightarrow0+}\operatorname{Im} F_\mu(x+i\varepsilon)= f(x)$$

c. There exists a measurable set $E\subset\mathbb{R}$ such that $\lambda(E)=0=\mu_s(\mathbb{R}\setminus E)$ such that $$ \lim_{\varepsilon\rightarrow0+}\operatorname{Im} F_\mu(x+i\varepsilon)= \infty,\qquad x\in E $$

d. For all $x_0\in\mathbb{R}$,

$$ \lim_{\varepsilon\rightarrow0+}\varepsilon\operatorname{Im} F_\mu(x_0+i\varepsilon)= \mu(\{x_0\}) $$

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I hope this helps answer your question, or at least provides you with good reference that discusses interesting results and historical notes about the Stieltjes transform.

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