Piecewise Continuous Function with One-Sided Limits is Darboux Integrable

Theorem
A piecewise continuous function $f$ defined on an interval $\left[{a \,.\,.\, b}\right]$ is Riemann integrable.

Proof
Since $f$ is piecewise continuous, 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.

By the piecewise continuity of $f$ for the case $n=1$, $f$ is continuous on $\left({a \,.\,.\, b}\right)$, and $\displaystyle \lim_{x \to a+} f\left({x}\right)$ and $\displaystyle \lim_{x \to b-} f\left({x}\right)$ 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 $n+1$.

Since $f$ is piecewise continuous, a subdivision $P$ exists, $P = \left\{{x_0, \ldots, x_{n+1}}\right\}$ of $\left[{a \,.\,.\, b}\right]$, $x_0 = a$ and $x_{n+1} = b$, so that $f$ is continuous on $\left({x_{i−1} \,.\,.\, x_i}\right)$ for every $i \in \left\{{1, \ldots, n+1}\right\}$.

We intend to prove that $f$ is piecewise continuous 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 $\displaystyle \lim_{x \to x_{i−1}+} f\left({x}\right)$ and $\displaystyle \lim_{x \to x_i-} f\left({x}\right)$ exist for every $i \in \left\{{1, \ldots, n}\right\}$.

Therefore, all the requirements for $f$ to be piecewise continuous on $\left[{x_0 \,.\,.\, x_n}\right]$ for the subdivision $\left\{{x_0, \ldots, x_n}\right\}$ of $\left[{x_0 \,.\,.\, x_n}\right]$ are satisfied.

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

Since $f$ is piecewise continuous on $\left[{a \,.\,.\, b}\right]$ with respect to the subdivision $P$, $f$ is continuous on $\left({x_n \,.\,.\, x_{n+1}}\right)$.

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

Therefore, all the requirements for $f$ to be piecewise continuous 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 on $\left[{x_n \,.\,.\, b}\right]$.

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

Since $f$ is piecewise continuous on $\left[{a \,.\,.\, x_n}\right]$, and the number of intervals $\left({x_{i−1} \,.\,.\, x_i}\right)$ defined from the subdivision $\left\{{x_0, \ldots, x_n}\right\}$ equals $n$, $f$ is integrable on $\left[{a \,.\,.\, x_n}\right]$ by the induction hypothesis.

Since $f$ is piecewise continuous on $\left[{x_n \,.\,.\, b}\right]$, and the number of intervals $\left({x_{i−1} \,.\,.\, x_i}\right)$ defined from the subdivision $\left\{{x_n, x_{n+1}}\right\}$ equals 1, $f$ is integrable on $\left[{x_n \,.\,.\, b}\right]$ by the fact that the theorem is true for the base case.

Hence, by Additivity with Respect to the Interval of Integration and the existence of the integrals $\displaystyle \int_a^{x_n} f \left(x\right) \ \mathrm d x$ and $\displaystyle \int_{x_n}^b f \left(x\right) \ \mathrm d x$, the left hand side of $\displaystyle \int_a^b f \left(x\right) \ \mathrm d x = \displaystyle \int_a^{x_n} f \left(x\right) \ \mathrm d x + \displaystyle \int_{x_n}^b f \left(x\right) \ \mathrm d x$ exists, which finishes the proof.