Using Fubini's Theorem in Contour Integrals proof I have a few questions regarding the following proof:
Suppose that $\mathcal{A}$ is a unital Banach algebra, and that $g$ is a complex-valued function which is analytic on $\sigma(a)$ while $f$ is a complex-valued function which is analytic on $g(\sigma(a))$. Then $(f \circ g)(a) = f(g(a))$. 
Proof: Let $V$ be an open neighbourhood of $g(\sigma(a))$ upon which $f$ is analytic and consider $U = g^{-1}(V)$, an open neighbourhood of $\sigma(a)$. Let $\Gamma_{1}$ be a system of closed contours in $U$ such that 
\begin{align}
 &\text{Ind}_{\Gamma_{1}}(\lambda) = 1~~\text{for all } \lambda \in \sigma(a)~~\text{and}~~\text{Ind}_{\Gamma_{1}}(\lambda) \neq 0~~\text{implies that }\lambda \in U.
\end{align}
Let $\Gamma_{2}$ be a system of closed contours in $V$ such that
\begin{align}
\text{Ind}_{\Gamma_{2}}(\beta) = 1 \text{ for all }\beta \in g(\sigma(a))~~\text{and }\text{Ind}_{\Gamma_{2}}(\beta)=1~~\text{for all }\beta \in g(\Gamma_{1}).
\end{align}
Then we get 
\begin{align}
(f \circ g)(a) &= \frac{1}{2 \pi i} \int_{\Gamma_{1}}(f \circ g)(z)(z-a)^{-1}dz
\\& = \frac{1}{2 \pi i}\int_{\Gamma_{1}}f(g(z))(z-a)^{-1}dz
\\& = \frac{1}{2 \pi i}\int_{\Gamma_{1}}\frac{1}{2 \pi i} \int_{\Gamma_{2}}f(w)(w-g(z))^{-1}dw(z-a)^{-1}dz
\\& = \frac{1}{2 \pi i} \int_{\Gamma_{2}}f(w) \frac{1}{2 \pi i} \int_{\Gamma_{1}}(w - g(z))^{-1}(z-a)^{-1}dzdw
\\& = \frac{1}{2 \pi i} \int_{\Gamma_{2}}f(w)(w-g(a))^{-1}dw
\\& = f(g(a)).
\end{align}
Questions:
In equation 3 of the proof I assume we are using Cauchy Integral Formula. In equation 4, it seems that Fubini's theorem is being used, but I don't see how, since we have a Banach space valued integral $\int_{\Gamma_{1}}...dz$ and a complex valued integral $\int_{\Gamma_{2}}...dw$. How is this step obtained? Does it not matter that the one integral maps to $\mathcal{A}$ and the other to $\mathbb{C}$. If possible could you expand on how this equation is obtained?
Thanks for any assistance. 
 A: One of the nice things about Functional Analysis is that you can generally reduce to the scalar case by applying a linear functional to everything, rearranging scalar integrals, and then pulling the functional back outside. Then, knowing that you have enough functionals to separate points allows you to remove the functional from both sides of the resulting equation. You do, however, need the existence of the integrals without the functionals, or to define the integrals by how they interact with functionals. But the theorems to interchange orders of integration can be reduced to corresponding scalar cases.
In full detail: The Banach algebra $\mathcal{A}$ is a Banach space and its dual space $\mathcal{A}^{\star}$ consists of all continuous linear functionals on $\mathcal{A}$. Suppose you have a continuous rectifiable curve $\gamma(t) : [0,1]\rightarrow\mathbb{C}$ and a continuous function $f : \Omega\subseteq\mathbb{C}\rightarrow \mathcal{A}$ on an open region $\Omega$ containing the image of the curve $\gamma$. Then you can define a Riemann-Stieljes integral
$$
           \int_{\gamma} f(z)dz = \lim_{\|\mathcal{P}\|\rightarrow 0} \sum_{j}f(\gamma(t_j^{\star}))\{\gamma(t_j)-\gamma(t_{j-1})\},
$$
where the partition $\mathcal{P}$ consists of subdivision points $0 = t_0 < t_1 < \cdots < t_n = 1$ and intermediate points $t_j^{\star} \in [t_{j-1},t_j]$. The right side converges in the norm of $\mathcal{A}$ as $\|\mathcal{P}\|\rightarrow 0$. If $\Phi \in \mathcal{A}^{\star}$, then $\Phi$ is continuous and linear, which gives the existence of the following limits, independent of the choice of $\Phi$:
\begin{align}
         \Phi\left(\int_{\gamma}f(z)dz\right)& =\Phi\left(\lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{j}f(\gamma(t_j^{\star}))\{\gamma(t_j)-\gamma(t_{j-1})\}\right) \\
   & = \lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{j}\Phi(f(t_j^{\star}))\{\gamma(t_j)-\gamma(t_{j-1})\} \\
   & = \int_{\gamma}\Phi(f(z))dz
\end{align}
The final integral on the right exists because the vector integral on the far left exists, and the two are related as shown above. You can easily throw in a scalar function $\alpha(z)$, too:
$$
          \Phi\left(\int_{\gamma}\alpha(z)f(z)dz\right)=\int_{\gamma}\alpha(z)\Phi(f(z))dz.
$$
This becomes a fundamental building block for dealing with integrals.
Next, in your case, you have integrals
$$
       I=\int_{\Gamma_1}\left(\int_{\Gamma_2}f(w)(w-g(z))^{-1}dw\right)(z-a)^{-1}dz.
$$
These are defined as iterated integrals. The inner integrand is a continuous function of $w$ for fixed $z \in \Gamma_1$, and the inner integral gives a continuous function of $z$. So you can then multiply the scalar value of this inner integral by $(z-a)^{-1}$ and integrate with respect to $z$. In fact, by continuity of Banach algebra multiplication, you can also write $I$ as
$$
           I = \int_{\Gamma_1}\left(\int_{\Gamma_2}f(w)(w-g(z))^{-1}(z-a)^{-1}dw\right)dz.
$$
Now apply a linear functional $\Phi\in\mathcal{A}^{\star}$ and apply our identity twice:
\begin{align}
       \Phi(I) & =\int_{\Gamma_1}\Phi\left(\int_{\Gamma_2}f(w)(w-g(z))^{-1}(z-a)^{1}dw\right)dz \\
         & = \int_{\Gamma_1}\left(\int_{\Gamma_2}f(w)\Phi((w-g(z))^{-1}(z-a)^{-1})dw\right)dz
\end{align}
This reduces everything to scalar integrals. The function $\Phi(\cdots)$ has all of the properties of the vector function in terms of joint continuity, etc., which then allows you to interchange orders of integration and to apply our identity in reverse:
\begin{align}
   \Phi(I) & = \int_{\Gamma_2}\int_{\Gamma_1}f(w)\Phi((w-g(z))^{-1}(z-a)^{-1})dzdw \\
& = \int_{\Gamma_2}f(w)\int_{\Gamma_1}\Phi((w-g(z))^{-1}(z-a)^{-1})dzdw \\
     & = \int_{\Gamma_2}f(w)\Phi\left(\int_{\Gamma_1}(w-g(z))^{-1}(z-a)^{-1}dz\right)dw \\
     & = \Phi\left(\int_{\Gamma_2}f(w)\int_{\Gamma_1}(w-g(z))^{-1}(z-a)^{-1}dzdw\right)
\end{align}
Because this is true for all linear functional $\Phi \in \mathcal{A}^{\star}$, then
$$
        I = \int_{\Gamma_2}f(w)\int_{\Gamma_1}(w-g(z))^{-1}(z-a)^{-1}dzdw
$$
And that is exactly what you want, where the end result is that you can interchange orders of integration for the problem at hand.
