1. for every \(z\) in \(\Omega,\) \(x \in X \mapsto f(z, x)\) is \(\mu\)-measurable,

2. for any \(z_0 \in \Omega,\) there is a neighborhood \(V\) of \(z_0\) in \(\Omega\) and a \(\mu\)-integrable function \(g: X \to \mathbb{R}_+\) such that \[\forall \, z \in V, \;|f(z,x)| \leq g(x) \; \mbox{$\mu$-a.e.}\]

3. for \(\mu\)-almost every \(x \in X,\) the function \(z \in \Omega \mapsto f(z, x)\) is holomorphic.

Then the function \(z \in \Omega \mapsto \int_X f(z, x) \, d\mu(x)\) is holomorphic and its derivative at any order \(n\) is \[ \frac{\partial^n}{\partial z^n} \left[ \int_X f(z, x) \, d\mu(x) \right] = \int_X \partial_z^n f(z, x) \, d\mu(x). \]

The difference quotient of \(z \mapsto \int_X f(z, x) \, d\mu(x)\) at \(z_0\) is equal to \[ \int_X \frac{f(z_0+h, x) - f(z_0, x)}{h}\, d\mu(x). \] Let \(h\) be a complex number such that \(|h|< r/2.\) For \(\mu\)-almost every \(x \in X,\) the function \(\phi: t \in [0,1] \mapsto f(z_0 + t h, x)\) is continuous on \([0,1],\) differentiable on \(\left]0,1\right[\) and satisfies \[ |\phi'(t)| = |\partial_z f(z_0+th, x)| |h| \leq \frac{g(x)}{r} |h|. \] Hence, the mean value inequality yields \[ \left| \frac{f(z_0+h, x) - f(z_0, x)}{h} \right| = \frac{|\phi(1) - \phi(0)|}{|h|} \leq \frac{4g(x)}{r}. \] Since \[ \lim_{h \to 0} \frac{f(z_0+h, x) - f(z_0, x)}{h} = \partial_z f(z_0, x) \; \mbox{ $\mu$-a.e.}, \] Lebesgue’s dominated convergence theorem provides the result for \(n=1.\) Now, the function \(\partial_z f\) also satisfies the three assumptions required by the theorem, hence by induction, the theorem statement holds at any order \(n.\) ‌\(\blacksquare\)

1. \(f\) is a continuous function.

2. for any \(w \in \Lambda,\) the function \(z \in \Omega \mapsto f(z, w)\) is holomorphic.

Then, for any sequence of rectifiable paths \(\gamma\) of \(\Lambda,\) the function \(z \in \Omega \mapsto \int_{\gamma} f(z, w) \, dw\) is holomorphic and \[ \frac{\partial}{\partial z} \left[ \int_{\gamma} f(z, w) \, dw \right] = \int_{\gamma} \partial_z f(z, w) \, dw. \]

1. For any \(z \in \Omega,\) the function \(t \in [0,1] \mapsto f(z, \gamma(t)) \gamma'(t)\) is continuous and therefore Lebesgue measurable.

2. Let \(z_0 \in \Omega\) and let \(r>0\) be such that \(K = \overline{D}(z_0, r) \subset \Omega.\) The restriction of \(f\) to the compact set \(K \times \gamma([0,1])\) is bounded by some constant \(\kappa.\) Therefore, for any \(z \in D(z_0,r),\) the function \(t \in [0,1] \mapsto f(z, \gamma(t)) \gamma'(t)\) is dominated by \(t \in [0,1] \mapsto \kappa |\gamma'(t)|\) which is Lebesgue integrable.

3. For any \(t \in [0,1],\) the function \(z \in \Omega \mapsto f(z, \gamma(t)) \gamma'(t)\) is holomorphic; its derivative is \(\partial_z f(z, \gamma(t)) \gamma'(t).\)

Consequently, the differentiation of Lebesgue integrals theorem provides the existence of \(\partial_z\left[ \int_{\gamma} f(z, w) \, dw \right]\) and its value: \[ \frac{\partial}{\partial z} \left[ \int_{\gamma} f(z, w) \, dw \right] = \int_{[0,1]} \partial_z f(z, \gamma(t)) \gamma'(t) \, dt. \] The right-hand side is equal to \(\int_{\gamma} \partial_z f(z, w) \, dw.\) ‌\(\blacksquare\)

1. For any \(s \in \Omega,\) the function \(t\mapsto f(t) e^{-st}\) is Lebesgue measurable.

2. Let \(s \in \Omega\) and let \(r > 0\) be such that \(\epsilon = \mathrm{Re}(s) - \sigma - r > 0.\) For any \(w \in D(s, r),\) we have \(\mathrm{Re}(w) > \mathrm{Re}(s) - r = \sigma + \epsilon,\) thus \[ \int_{\mathbb{R}_+} |f(t) e^{-wt}| \, dt = \int_{\mathbb{R}_+} |f(t)| e^{-\mathrm{Re}(w)t} \, dt \leq \int_{\mathbb{R}_+} |f(t)| e^{-(\sigma + \epsilon)t} \, dt < +\infty. \]

3. For almost any \(t \geq 0,\) \(s \mapsto f(t) e^{-st}\) is holomorphic and \[\partial_s [f(t) e^{-st}] = -t f(t) e^{-st}.\]

We can therefore differentiate under the integral sign and obtain \[ \frac{\partial}{\partial s}\int_0^{+\infty} f(t) e^{-st} \, dt = \int_0^{+\infty} -t f(t) e^{-st} \, dt = \mathcal{L} [t \mapsto -t f(t)](s) \] as expected. ‌\(\blacksquare\)

Now the function \[ z \in \Omega \mapsto \int_{\gamma} g(z, w) dw \] clearly extends the function of the lemma statement. It also satisfies the assumptions of the complex-differentiation of line integrals result, thus it is holomorphic. ‌\(\blacksquare\)

Introduce the holomorphic extension \(h\) to \(\Omega\) of
\[ z \in \Omega \setminus \gamma([0,1]) \mapsto \frac{1}{i2\pi} \int_{\gamma} \frac{f(z) - f(w)}{z - w} dw \] and define the function \(\phi: \mathbb{C} \mapsto \mathbb{C}\) by \[ \phi(z) = h(z) \; \mbox{ if } \, z \in \Omega, \; \phi(z) = -\frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w - z} \, dw \; \mbox{ if } \; z\in \mathrm{Ext} \, \gamma. \] This definition is unambiguous: if \(z \in \Omega \cap \mathrm{Ext} \, \gamma,\) then \[ \begin{split} h(z) &= \frac{1}{i2\pi} \int_{\gamma} \frac{f(z) - f(w)}{z - w} dw \\ &= f(z) \mathrm{ind}(\gamma, z) - \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{z - w} dw \\ &= - \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{z - w} dw \end{split}. \] The function \(\phi\) is holomorphic on \(\Omega\) and also on \(\mathrm{Ext}\, \gamma\) by the complex-differentiation of line integrals theorem. Hence, it is holomorphic on \(\mathbb{C}.\) Additionally, if \(|z|> r= \max \{|w| \; | \; w \in \gamma([0,1])\},\) then \(z \in \mathrm{Ext} \, \gamma,\) thus if \(M\) is an upper bound of \(f\) on the image of \(\gamma,\) \[ |\phi(z)| \leq \frac{1}{2\pi}\frac{M}{|z| - r} \times \ell(\gamma) \] and \(|\phi(z)| \to 0\) when \(|z|\to +\infty.\) By Liouville’s Theorem, \(\phi\) is identically zero; hence, if \(z\in\Omega,\) \[ \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{z - w} dw = \frac{1}{i2\pi} \int_{\gamma} \frac{f(z)}{z - w} dw = \mathrm{ind}(\gamma, z) f(z), \] which is Cauchy’s integral formula. ‌\(\blacksquare\)