Why is $f(z)$ holomorphic?

Question

Given a finite Borel measure $\mu$ on $\mathbb{R}$ we can define the function: $$f(z) = \int_{\mathbb{R}}\frac{d\mu(x)}{x-z}.$$ I need to show that $f$ is holomorphic on $\mathbb{C}^{+} =\{z \in \mathbb{C}: \operatorname{Im}(z) > 0\}$, but I don't know how to finish the argument. I was trying to use Morera's Theorem together with Fubini's Theorem to prove: $$\int_{\mathbb{C}^{+}} f(z)dz = \int_{\mathbb{R}}\int_{\mathbb{C}^{+}}\frac{dz}{x-z}d\mu(x) = 0$$ but I can't convince myself that: $$\int_{\mathbb{C}^{+}}\frac{dz}{x-z} = 0.$$ Is there any reason for this integral to be zero? Maybe expand $(x-z)^{-1}$ in power series of $z$ or something like this?

Answer 1

Fix $z\in\mathbb{C}^+$. Consider the map

$$g:h\mapsto\frac{\int_{\mathbb{R}}\frac{\mathrm{d}\mu(x)}{x-(z+h)}-\int_{\mathbb{R}}\frac{\mathrm{d}\mu(x)}{x-z}}{h}$$

for $h$ in some neighborhood of $0$ in $\mathbb{C}$. In particular, we choose a neighborhood small enough so that $z+h\in\mathbb{C}^+$. If we can show that $\lim_{h\to0}g(h)$ exists, then we are done. Now it is easy to check that

$$g(h)=\int_{\mathbb{R}}\frac{\mathrm{d}\mu(x)}{(x-z-h)(x-z)}.$$

It's now easy to see that $x\mapsto \frac{1}{(x-z-h)(x-z)}$ is bounded on $\mathbb{R}$, as $z,z+h\in\mathbb{C}^+$, and that we can find such a bound independent of $h$. Thus $\left\lvert \frac{1}{(x-z-h)(x-z)}\right\rvert$ is dominated by some constant function, which is clearly integrable as $\mu$ is finite. Choosing any sequence $\{h_j\}_{j\in\mathbb{N}}$ in the domain of $f$ with limit $0$, we then have that

$$\lim_{j\to\infty}g(h_j)=\lim_{j\to\infty}\int_{\mathbb{R}}\frac{\mathrm{d}\mu(x)}{(x-z-h_j)(x-z)}=\int_{\mathbb{R}}\frac{\mathrm{d}\mu(x)}{(x-z)^2}$$

by the dominated convergence theorem. As the sequence was arbitrary, the limit exists and equals the above (which is also the derivative of your function at $z$).

So as $z\in\mathbb{C}^+$ was arbitrary, the function is complex differentiable on all of $\mathbb{C}^+$, i.e. it's holomorphic on $\mathbb{C}^+$.