# 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:

$f$ is **piecewise continuous with one-sided limits** if and only if:

- there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that, for all $i \in \set {1, 2, \ldots, n}$:

- $(1): \quad f$ is continuous on $\openint {x_{i − 1} } {x_i}$

- $(2): \quad$ 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.

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

$\blacksquare$