# 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 $\left[{a \,.\,.\, b}\right]$.

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 $\displaystyle \lim_{x \mathop \to {x_{i − 1} }^+} \map f x$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \map f x$ exist.

Then:

for all $i \in \left\{ {1, 2, \ldots, n}\right\}$, $f$ is uniformly continuous on $\left({x_{i − 1} \,.\,.\, x_i}\right)$ .

## Proof

We have that $f$ is continuous on $\left({x_{i − 1} \,.\,.\, x_i}\right)$ for every $i \in \left\{ {1, 2, \ldots, n}\right\}$.

Since $f$ is piecewise continuous with one-sided limits, the one-sided limits:

$\displaystyle \lim_{x \mathop \to {x_{i−1} }^+} f \left({x}\right)$

and:

$\displaystyle \lim_{x \mathop \to {x_i}^-} f \left({x}\right)$

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

Thus, from Extendability Theorem for Function Continuous on Open Interval, 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 $\left({x_{i − 1} \,.\,.\, x_i}\right)$

for an arbitrary $i$ in $\left\{{1, 2, \ldots, n}\right\}$

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

We have that $\left({x_{i − 1} \,.\,.\, x_i}\right)$ is a subset of $\left[{x_{i − 1} \,.\,.\, x_i}\right]$.

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

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

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

$\blacksquare$