Continuous Complex Function is Complex Riemann Integrable

Theorem
Let $\closedint a b$ be a closed real interval.

Let $f: \closedint a b \to \C$ be a continuous complex function.

Then $f$ is complex Riemann integrable over $\closedint a b$.

Proof
Define the real function $x: \closedint a b \to \R$ by:


 * $\forall t \in \closedint a b : \map x t = \map \Re {\map f t}$

Define the real function $y: \closedint a b \to \R$ by:


 * $\forall t \in \closedint a b : \map y t = \map \Im {\map f t}$

where:
 * $\map \Re {\map f t}$ denotes the real part of the complex number $\map f t$
 * $\map \Im {\map f t}$ denotes the imaginary part of $\map f t$.

From Real and Imaginary Part Projections are Continuous, it follows that $\Re: \C \to \R$ and $\Im: \C \to \R$ are continuous functions.

From Continuity of Composite Mapping, it follows that $x$ and $y$ are continuous.

From Continuous Real Function is Darboux Integrable, it follows that $x$ and $y$ are Darboux integrable over $\closedint a b$.

By definition, it follows that $f$ is complex Riemann integrable.