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It is a known fact that Borel measures are uniquely determined by their Fourier transforms. This is the motivation for the following question.

I came across the concept of a Borel transform of a Borel measure. Suppose $\mathbb{R}$ is equipped with the usual Borel $\sigma$-algebra $\mathbb{B}_{\mathbb{R}}$ and let $\mu$ be a Borel measure. Letting: $$\mathbb{C}_{+} := \{z \in \mathbb{C}: \operatorname{Im}z > 0\},$$ the Borel transform of $\mu$ is defined as the function $F: \mathbb{C}_{+} \to \mathbb{C}_{+}$ given by: $$F(z) = \int_{\mathbb{R}}\frac{1}{x-z}d\mu(x) $$

So, my question is: do Borel transforms uniquely determine a Borel measure? In other words, if: $$\int_{\mathbb{R}}\frac{1}{x-z}d\mu(x) = \int_{\mathbb{R}}\frac{1}{x-z}d\nu(x)$$ holds for every $z \in \mathbb{C}_{+}$ for some other Borel measure $\nu$ on $\mathbb{R}$, does it follow that $\mu = \nu$?

This is a new concept for me, so that this might be a standard result. However, I have looked for this result on the internet, and found nothing about it.

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    $\begingroup$ Yes, it is indeed a standard result. The measure is determined by $F$ and in fact it can be recovered as $d\mu(x) = \lim_{y\to 0+}(1/\pi)\textrm{Im }F(x+iy)\, dx$ (in weak $*$ sense). $\endgroup$
    – Christian Remling
    Commented Apr 13, 2023 at 0:42
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    $\begingroup$ This is discussed in many spectral theory books, for example Section 3.4 here: mat.univie.ac.at/~gerald/ftp/book-schroe/index.html $\endgroup$
    – Christian Remling
    Commented Apr 13, 2023 at 0:45
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    $\begingroup$ You need to impose a growth condition at infty on your measure to ensure that "Borel transform" exists. (A more common name for this is Cauchy transform, btw). $\endgroup$
    – Alexandre Eremenko
    Commented Apr 13, 2023 at 1:45

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