Piecewise Continuous Function with One-Sided Limits is Uniformly Continuous on Each Piece

Theorem
Let a real function $f$ defined on an interval $\left[{a \,.\,.\, b}\right]$ be piecewise continuous.

Since $f$ is piecewise continuous, a finite subdivision $\left\{{x_0, \ldots, x_n}\right\}$ of $\left[{a \,.\,.\, b}\right]$, $x_0 = a$ and $x_n = b$, exists so that $f$ is continuous on $\left({x_{i−1} \,.\,.\, x_i}\right)$ for every $i \in \left\{{1, \ldots, n}\right\}$.

Then $f$ is uniformly continuous on $\left({x_{i−1} \,.\,.\, x_i}\right)$ for every $i \in \left\{{1, \ldots, n}\right\}$.

Proof
We know that $f$ is continuous on $\left({x_{i−1} \,.\,.\, x_i}\right)$ for every $i \in \left\{{1, \ldots, n}\right\}$.

Since $f$ is piecewise continuous, the one-sided limits $\displaystyle \lim_{x \to x_{i−1}^+} f\left({x}\right)$ and $\displaystyle \lim_{x \to x_i^-} f\left({x}\right)$ exist for every $i \in \left\{{1, \ldots, n}\right\}$.

These facts ensure by Extendability Theorem for Functions Continuous on Open Intervals that, for an arbitrary $i$ in $\left\{{1, \ldots, n}\right\}$, a function $f_i$ exists that satisfies:


 * $f_i$ is defined on $\left[{x_{i−1} \,.\,.\, x_i}\right]$.


 * $f_i$ is continuous on $\left[{x_{i−1} \,.\,.\, x_i}\right]$.


 * $f_i$ equals $f$ on $\left({x_{i−1} \,.\,.\, x_i}\right)$.

Continuous Function on Closed Interval is Uniformly Continuous ensures that $f_i$ is uniformly continuous on $\left[{x_{i−1} \,.\,.\, x_i}\right]$.

Since $\left({x_{i−1} \,.\,.\, x_i}\right)$ is a subset of $\left[{x_{i−1} \,.\,.\, x_i}\right]$, $f_i$ is uniformly continuous on $\left({x_{i−1} \,.\,.\, x_i}\right)$.

Since $f_i$ equals $f$ on $\left({x_{i−1} \,.\,.\, x_i}\right)$, $f$ too is uniformly continuous on $\left({x_{i−1} \,.\,.\, x_i}\right)$.

Since $i$ is arbitrary, this results holds for all $i \in \left\{{1, \ldots, n}\right\}$.