Continuous Complex Function is Complex Riemann Integrable
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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 Composite of Continuous Mappings is Continuous, 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.
$\blacksquare$
Sources
- 2001: Christian Berg: Kompleks funktionsteori $\S 2.1$