Piecewise Continuous Function with One-Sided Limits is Darboux Integrable

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.

Then $f$ is Riemann integrable.

Proof
We have that $f$ is piecewise continuous with one-sided limits on $\left[{a \,.\,.\, b}\right]$.

Therefore, there exists a finite subdivision $\left\{{x_0, \ldots, x_n}\right\}$ of $\left[{a \,.\,.\, b}\right]$, $x_0 = a$ and $x_n = b$, such that $f$ is continuous on $\left({x_{i - 1} \,.\,.\, x_i}\right)$ for every $i \in \left\{{1, \ldots, n}\right\}$.

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

We shall use proof by induction on the number of intervals.

We start with the case $n = 1$, the base case.

We need to prove that the theorem is true for this case.

Piecewise continuity with one-sided limits of $f$ for the case $n = 1$ means that:
 * $f$ is continuous on $\left({a \,.\,.\, b}\right)$
 * $\displaystyle \lim_{x \mathop \to a^+} f \left({x}\right)$ and $\displaystyle \lim_{x \mathop \to b^-} f \left({x}\right)$ exist.

By Integrability Theorem for Functions Continuous on Open Intervals, $f$ is Riemann integrable on $\left[{a \,.\,.\, b}\right]$.

This finishes the proof for the case $n = 1$.

Assume that the theorem is true for some $n$, $n \geq 1$.

We need to prove that the theorem is true for the case $n + 1$.

Piecewise continuity with one-sided limits of $f$ for the case $n + 1$ gives that:
 * there exists a subdivision $P$, $P = \left\{{x_0, \ldots, x_{n + 1}}\right\}$ of $\left[{a \,.\,.\, b}\right]$, $x_0 = a$ and $x_{n + 1} = b$
 * $f$ is continuous on $\left({x_{i - 1} \,.\,.\, x_i}\right)$ for every $i \in \left\{{1, \ldots, n + 1}\right\}$.

First we show that $f$ is piecewise continuous with one-sided limits on $\left[{a \,.\,.\, x_n}\right]$ and $\left[{x_n \,.\,.\, b}\right]$.

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

Also, 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 $i \in \left\{{1, \ldots, n}\right\}$.

Therefore, all the requirements for $f$ to be piecewise continuous with one-sided limits on $\left[{x_0 \,.\,.\, x_n}\right]$ for the subdivision $P \setminus \left\{x_{n + 1}\right\}$ of $\left[{x_0 \,.\,.\, x_n}\right]$ are satisfied.

As $x_0 = a$, we have shown that $f$ is piecewise continuous with one-sided limits on $\left[{a \,.\,.\, x_n}\right]$.

We have that $f$ is piecewise continuous with one-sided limits on $\left[{a \,.\,.\, b}\right]$ with respect to the subdivision $P$.

Therefore, $f$ is continuous on $\left({x_n \,.\,.\, x_{n + 1}}\right)$.

Also, the one-sided limits $\displaystyle \lim_{x \mathop \to x_n^+} f \left({x}\right)$ and $\displaystyle \lim_{x \mathop \to x_{n + 1}^-} f \left({x}\right)$ exist.

Therefore, all the requirements for $f$ to be piecewise continuous with one-sided limits on $\left[{x_n \,.\,.\, x_{n + 1}}\right]$ for the subdivision $\left\{{x_n, x_{n + 1}}\right\}$ of $\left[{x_n \,.\,.\, x_{n + 1}}\right]$ are satisfied.

As $x_{n + 1} = b$, we have shown that $f$ is piecewise continuous with one-sided limits on $\left[{x_n \,.\,.\, b}\right]$.

Next we show that $f$ is integrable on $\left[{a \,.\,.\, x_n}\right]$ and $\left[{x_n \,.\,.\, b}\right]$:

We have that $f$ is piecewise continuous with one-sided limits on $\left[{a \,.\,.\, x_n}\right]$.

The number of intervals $\left({x_{i - 1} \,.\,.\, x_i}\right)$ defined from the subdivision $\left\{{x_0, \ldots, x_n}\right\}$ equals $n$.

Therefore, $f$ is integrable on $\left[{a \,.\,.\, x_n}\right]$ by the induction hypothesis.

We have that $f$ is piecewise continuous with one-sided limits on $\left[{x_n \,.\,.\, b}\right]$.

The number of intervals $\left({x_{i - 1} \,.\,.\, x_i}\right)$ defined from the subdivision $\left\{{x_n, x_{n + 1}}\right\}$ equals 1.

Therefore, $f$ is integrable on $\left[{x_n \,.\,.\, b}\right]$ by the fact that the theorem is true for the base case.

We have that $f$ is integrable on $\left[{a \,.\,.\, x_n}\right]$ and $\left[{x_n \,.\,.\, b}\right]$.

Therefore, $f$ is integrable on $\left[{a \,.\,.\, x_n}\right] \cup \left[{x_n \,.\,.\, b}\right]$ by Existence of Integral on Union of Adjacent Intervals.

Accordingly, $f$ is integrable on $\left[{a \,.\,.\, b}\right]$ as $\left[{a \,.\,.\, b}\right] = \left[{a \,.\,.\, x_n}\right] \cup \left[{x_n \,.\,.\, b}\right]$.