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

Theorem
Let $f$ be a real function defined on a closed interval $\closedint a b$.

Let $f$ be piecewise continuous with one-sided limits:

Then:
 * for all $i \in \set {1, 2, \ldots, n}$, $f$ is uniformly continuous on $\openint {x_{i − 1} } {x_i}$.

Proof
We have that $f$ is continuous on $\openint {x_{i − 1} } {x_i}$ for every $i \in \set {1, 2, \ldots, n}$.

Since $f$ is piecewise continuous with one-sided limits, the one-sided limits:
 * $\ds \lim_{x \mathop \to {x_{i−1} }^+} \map f x$

and:
 * $\ds \lim_{x \mathop \to {x_i}^-} \map f x$

exist for every $i \in \set {1, 2, \ldots, n}$.

Thus, from Extendability Theorem for Function Continuous on Open Interval, a function $f_i$ exists that satisfies:


 * $f_i$ is defined on $\closedint {x_{i − 1} } {x_i}$


 * $f_i$ is continuous on $\closedint {x_{i − 1} } {x_i}$


 * $f_i$ equals $f$ on $\openint {x_{i − 1} } {x_i}$

for an arbitrary $i$ in $\set {1, 2, \ldots, n}$.

From Continuous Function on Closed Real Interval is Uniformly Continuous, $f_i$ is uniformly continuous on $\closedint {x_{i − 1} } {x_i}$.

We have that $\openint {x_{i − 1} } {x_i}$ is a subset of $\closedint {x_{i − 1} } {x_i}$.

Hence $f_i$ is uniformly continuous on $\openint {x_{i − 1} } {x_i}$.

As $f_i$ equals $f$ on $\openint {x_{i − 1} } {x_i}$, $f$ also is uniformly continuous on $\openint {x_{i − 1} } {x_i}$.

Since $i$ is arbitrary, this results holds for all $i \in \set {1, 2, \ldots, n}$.