Piecewise Continuous Function with One-Sided Limits is Darboux Integrable

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Theorem

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

Let $f$ be piecewise continuous with one-sided limits on $\closedint a b$.


Then $f$ is Darboux integrable on $\closedint a b$.


Proof 1

We are given that $f$ is piecewise continuous with one-sided limits on $\closedint a b$.

From Piecewise Continuous Function with One-Sided Limits is Bounded, $f$ is a bounded piecewise continuous function.

The result follows from Bounded Piecewise Continuous Function is Darboux Integrable.

$\blacksquare$


Proof 2

We are given that $f$ is piecewise continuous with one-sided limits on $\closedint a b$.

Therefore, 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 $\openint {x_{i - 1} } {x_i}$
$\ds \lim_{x \mathop \to {x_{i - 1} }^+} \map f x$ and $\ds \lim_{x \mathop \to {x_i}^-} \map f x$ exist.


Note that $n$ is the number of intervals $\openint {x_{i - 1} } {x_i}$ defined from the (finite) subdivision $\set {x_0, x_1, \ldots, x_n}$.

We shall use proof by induction on the intervals.


For all $k \in \set {1, 2, \ldots, n}$, let $\map P k$ be the proposition:

$f$ is Darboux integrable on $\closedint {x_0} {x_k}$.


Basis for the Induction

$\map P 1$ is the case:

$f$ is Darboux integrable on $\closedint {x_{i - 1} } {x_i}$

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


Piecewise continuity with one-sided limits of $f$ for the case $n = 1$ means that:

$f$ is continuous on $\openint {x_{i - 1} } {x_i}$
$\ds \lim_{x \mathop \to {x_{i - 1} }^+} \map f x$ and $\ds \lim_{x \mathop \to {x_i}^-} \map f x$ exist.

By Integrability Theorem for Functions Continuous on Open Intervals, $f$ is Darboux integrable on $\closedint {x_{i - 1} } {x_i}$.

Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

$f$ is Darboux integrable on $\openint {x_0} {x_k}$.

from which it is to be shown that:

$f$ is Darboux integrable on $\openint {x_0} {x_{k + 1} }$.


Induction Step

This is the induction step:


By definition of a piecewise continuous function with one-sided limits, for every $i \in \set {1, 2, \ldots, k, k + 1}$:

$f$ is continuous on $\openint {x_{i - 1} } {x_i}$
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.


By the induction hypothesis, $f$ is Darboux integrable on $\closedint {x_0} {x_k}$.


From the basis for the induction, $f$ is Darboux integrable on $\closedint {x_k} {x_{k + 1} }$.


We have that $f$ is Darboux integrable on $\closedint {x_0} {x_k}$ and $\closedint {x_k} {x_{k + 1} }$.

Therefore, by Existence of Integral on Union of Adjacent Intervals:

$f$ is Darboux integrable on $\closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$.

We have that:

$\closedint {x_0} {x_{k + 1} } = \closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$

Accordingly, $f$ is Darboux integrable on $\closedint {x_0} {x_{k + 1} }$.


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$f$ is Darboux integrable on $\closedint a b$.

$\blacksquare$


Sources

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