# Definition:Piecewise Continuous Function

## Definition

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

$f$ is piecewise continuous 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 $i \in \set {1, 2, \ldots, n}$, $f$ is continuous on $\openint {x_{i − 1} } {x_i}$.

Further conditions can be placed on this definition to add specialisation as necessary:

### With Improper Integrals

$f$ is piecewise continuous with improper integrals 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 improper integrals $\ds \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ all exist.

### Bounded

$f$ is a bounded piecewise continuous function 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:
$(1): \quad$ for all $i \in \set {1, 2, \ldots, n}$, $f$ is continuous on $\openint {x_{i − 1} } {x_i}$
$(2): \quad$ $f$ is bounded on $\closedint a b$.

### 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.

## Also known as

Some sources hyphenate: piecewise-continuous.

## Also defined as

There are other definitions of Piecewise Continuous Function.

For example, the following variants exist:

### Variant 1: $f$ is piecewise defined

Let $\closedint a b$ be a closed interval.

Let $\set {x_0, x_1, \ldots, x_n}$ be a finite subdivision of $\closedint a b$, where $x_0 = a$ and $x_n = b$.

Let $f$ be a real function defined on $\closedint a b \setminus \set {x_0, x_1, \ldots, x_n}$.

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

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.

### Variant 2: $f$ is complex-valued

Let $f$ be a complex-valued function defined on a closed interval $\closedint a b$.

$f$ is piecewise continuous 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 $i \in \set {1, 2, \ldots, n}$, $f$ is continuous on $\openint {x_{i − 1} } {x_i}$.

### Variant 3: $f$ has unbounded domain

Let $f$ be a real function defined on $\R$.

$f$ is piecewise continuous if and only if:

for any closed interval $\closedint a b$:
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 $i \in \set {1, 2, \ldots, n}$, $f$ is continuous on $\openint {x_{i − 1} } {x_i}$.