# Integrability Theorem for Functions Continuous on Open Intervals

## Theorem

Let $f$ be a real function defined on a closed interval $\closedint a b$ such that $a < b$.

Let $f$ be continuous on $\openint a b$.

Let the one-sided limits $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist.

Then $f$ is Riemann integrable on $\closedint a b$.

## Proof

We start by showing that $f$ is bounded.

A real function $g$ defined on $\closedint a b$ exists by Extendability Theorem for Function Continuous on Open Interval such that:

- $g$ equals $f$ on $\openint a b$
- $g$ is continuous on $\closedint a b$

Therefore:

- $g$ is bounded on $\closedint a b$ by Continuous Function on Compact Subspace of Euclidean Space is Bounded and Closed Real Interval is Compact

Accordingly, $g$ is bounded on $\openint a b$ as $\openint a b$ is a subset of $\closedint a b$.

Therefore, $f$ is bounded on $\openint a b$ as $f$ equals $g$ on $\openint a b$.

Also, $f$ is bounded on the set $\set {a, b}$ by the maximum $\map {\max} {\size {\map f a}, \size {\map f b}}$.

Therefore:

- $f$ is bounded on $\closedint a b$

We have:

- $f$ is bounded on $\closedint a b$
- $f$ is continuous on $\openint a b$

Therefore:

- $f$ is Riemann integrable on $\closedint a b$ by Bounded Function Continuous on Open Interval is Riemann Integrable

$\blacksquare$