Piecewise Continuous Function with One-Sided Limits is Darboux Integrable/Proof 2
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
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 $\closedint a b$.
$\blacksquare$