Continuous Complex Function is Complex Riemann Integrable

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Theorem

Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval.

Let $f: \left[{a \,.\,.\, b}\right] \to \C$ be a continuous complex function.


Then $f$ is complex Riemann integrable over $\left[{a \,.\,.\, b}\right]$.


Proof

Define the real function $x: \left[{a \,.\,.\, b}\right] \to \R$ by:

$\forall t \in \left[{a \,.\,.\, b}\right] : x \left({t}\right) = \operatorname{Re} \left({f \left({t}\right) }\right)$

Define the real function $y: \left[{a \,.\,.\, b}\right] \to \R$ by:

$\forall t \in \left[{a \,.\,.\, b}\right] : y \left({t}\right) = \operatorname{Im} \left({f \left({t}\right) }\right)$

Here, $\operatorname{Re} \left({f \left({t}\right) }\right)$ denotes the real part of the complex number $f \left({ t }\right)$, and $\operatorname{Im} \left({f \left({t}\right) }\right)$ denotes the imaginary part of $f \left({t}\right)$.


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

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

From Continuous Function is Riemann Integrable, it follows that $x$ and $y$ are Riemann integrable over $\left[{a \,.\,.\, b}\right]$.

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

$\blacksquare$


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