# Equvalent definitions of complex integrals over real intervals

My textbook in complex analysis defines a complex integral in the following way

Let $a, b \in \mathbb R$ and $g : [a, b] \to \mathbb C$ be a continuous complex-valued function. If $u$ and $v$ are both real-valued Riemann integratable functions over $[a, b]$ and $g(z) = u(z) + iv(z)$, then we define

$\displaystyle\int_a^b g(t) \; \mathrm dt := \displaystyle\int_a^b u(t) \; \mathrm dt + i \displaystyle\int_a^b v(t) \; \mathrm dt$

Let now make an alternative definition

Let $[a, b]$ be partitioned into $[a, b] = \displaystyle \bigcup_{k=1}^n \left[x_k, x_{k+1}\right]$ with $\begin{cases}x_1 = a \\ x_{k+1} > x_k \\ x_{n+1} = b\end{cases}$

Let $\Psi$ and $\Phi$ be complex valued functions such that

$\begin{cases}\mathcal{Re}(\Psi(z)) \leq \mathcal{Re}(f(z)) \leq \mathcal{Re}(\Phi(z)) \\ \mathcal{Im}(\Psi(z)) \leq \mathcal{Im}(f(z)) \leq \mathcal{Im}(\Phi(z)) \\ \Psi \text{ and } \Phi \text{ constant on } (x_k, x_{k+1}) \end{cases}$

Define the integral of $\Psi$ and $\Phi$ over $[a, b]$, denoted $I(\Psi)$ and $I(\Phi)$, as

$I(\Psi) := \displaystyle\sum_{k=1}^n \Psi(\xi_k) (x_{k+1}-x_k)$ where $\xi_k \in (x_k, x_{k+1})$

$I(\Phi) := \displaystyle\sum_{k=1}^n \Phi(\xi_k) (x_{k+1}-x_k)$ where $\xi_k \in (x_k, x_{k+1})$

We say that $g$ is Riemann integratable over $[a, b]$ if and only if $\forall \varepsilon > 0 \; \exists \Psi, \Phi : |I(\Phi) - I(\Psi)| < \varepsilon$

The unique complex constant $z_0 \in \mathbb C$ satisfying $\begin{cases}\mathcal{Re}(I(\Psi)) \leq \mathcal{Re}(z_0) \leq \mathcal{Re}(I(\Phi)) \\ \mathcal{Im}(I(\Psi)) \leq \mathcal{Im}(z_0) \leq \mathcal{Im}(I(\Phi)) \end{cases}$ for all functions $\Psi$ and $\Phi$ is called the integral of $g$ over $[a, b]$. We denote this number $\displaystyle\int_a^b g(t) \; \mathrm dt$.

Are these definitions equivalent? Can one extend the definition(s) above to compact sets $G \subseteq \mathbb C$ where $\partial G \subseteq G$ is a null set?

• Your second part gives a definition of the Riemann-Darboux integral. That $f$ is complex valued doesn't change anything. Generalizing to an integral over $\mathbb{R}^n$ is straighforward. In general "complex integration" refers to something else, where we use that $\mathbb{C}$ is a field so we don't identify it to $\mathbb{R}^2$ (see holomorphic functions and contour integration) – reuns Aug 30 '17 at 19:19
• @reuns As I understand it, things in $\mathbb C$ can be very different from $\mathbb R^2$. But the extension to integration over compact subsets to $\mathbb C$ (or even $\mathbb C^n$) is "as you would expect" from real multi-variable analysis? – Markus Klyver Aug 30 '17 at 19:27
• From $f : \mathbb{C} \supset D \to \mathbb{C}$ continuous and $D = \{ z \in \mathbb{C}, |z| \le 1\}$ the closed unit disk, if you identify $\mathbb{C}$ with $\mathbb{R}^2$ then you can define $\iint_D f(x_1+i x_2) d^2 x$. That $f$ is complex-valued doesn't change anything (linearity of $\int$). This integral over $\mathbb{R}^2$ will be defined by replacing $\sup_{x \in [x(i),x(i+1)]} f(x) (x(i+1)-x(i))$ by $\sup_{x \in [x_1(i),x_1(i+1)] \times [x_2(j),x_2(j+1)]} f(x) (x_1(i+1)-x_1(i))(x_2(j+1)-x_2(j))$ in Darboux sums – reuns Aug 30 '17 at 19:34
• Note that the Riemann integral for real-valued functions can be introduced in several equivalent ways. In any case, your "alternative definition" is nothing else but an unpacking of your first definition. – Christian Blatter Aug 31 '17 at 7:15