Bounded Piecewise Continuous Function has Improper Integrals

Theorem
Let $f$ be a real function defined on a closed interval $\left[{a \,.\,.\, b}\right]$, $a < b$.

Let $f$ be piecewise continuous and bounded on $\left[{a \,.\,.\, b}\right]$.

Then $f$ is a piecewise continuous function with improper integrals.

Proof
Let $f$ be piecewise continuous and bounded on $\left[{a \,.\,.\, b}\right]$.

It is sufficient to prove that the improper integral $\displaystyle \int_{x_{i - 1}+}^{x_i-} f \left({x}\right) \rd x$ exists for every $i \in \left\{{1, 2, \ldots, n}\right\}$.

Let $i \in \left\{{1, 2, \ldots, n}\right\}$.

Let $c$ be a point in $\left({x_{i − 1} \,.\,.\, x_i}\right)$.

By definition, the improper integral $\displaystyle \int_{x_{i - 1}+}^{x_i-} f \left({x}\right) \rd x$ exists :
 * $\displaystyle \lim_{\gamma \mathop \to x_{i - 1}+} \int_\gamma^c f \left({x}\right) \rd x$

and:
 * $\displaystyle \lim_{\gamma \mathop \to x_i-} \int_c^\gamma f \left({x}\right) \rd x$

both exist.

By Bounded Piecewise Continuous Function is Riemann Integrable we know that $f$ is integrable on $\left[{a \,.\,.\, b}\right]$.

Therefore, $f$ is integrable on every closed subinterval of $\left[{a \,.\,.\, b}\right]$.

Accordingly, the following definite integrals:
 * $\displaystyle \int_c^{x_i} f \left({x}\right) \rd x$
 * $\displaystyle \int_c^\gamma f \left({x}\right) \rd x$
 * $\displaystyle \int_\gamma^c f \left({x}\right) \rd x$
 * $\displaystyle \int_{x_{i-1}}^c f \left({x}\right) \rd x$

all exist.

Note that $f$ is bounded on $\left[{a \,.\,.\, b}\right]$ as $f$ is piecewise continuous and bounded.

Therefore, a bound $B$ exists for $f$ on $\left[{a \,.\,.\, b}\right]$.

We have:

which approaches $0$ as $\gamma$ approaches $x_i$.

This shows that $\displaystyle \lim_{\gamma \mathop \to x_i-} \int_c^\gamma f \left({x}\right) \rd x$ equals $\displaystyle \int_c^{x_i} f \left({x}\right) \rd x$.

From this we gather that $\displaystyle \lim_{\gamma \mathop \to x_i-} \int_c^\gamma f \left({x}\right) \rd x$ exists.

Also:

which approaches $0$ as $\gamma$ approaches $x_{i − 1}$.

This shows that $\displaystyle \lim_{\gamma \mathop \to x_{i - 1}+} \int_\gamma^c f \left({x}\right) \rd x$ equals $\displaystyle \int_{x_{i - 1} }^c f \left({x}\right) \rd x$.

From this we gather that $\displaystyle \lim_{\gamma \mathop \to x_{i - 1}+} \int_\gamma^c f \left({x}\right) \rd x$ exists.

Since $i$ is arbitrary, we have shown that $\displaystyle \lim_{\gamma \mathop \to x_i-} \int_c^\gamma f \left({x}\right) \rd x$ and $\displaystyle \lim_{\gamma \mathop \to x_{i-1}+} \int_\gamma^c f \left({x}\right) \rd x$ exist for every $i \in \left\{{1, \ldots, n}\right\}$.

Therefore, $\displaystyle \int_{x_{i-1}+}^{x_i-} f \left({x}\right) \rd x$ exists for every $i \in \left\{{1, \ldots, n}\right\}$.

That is, $f$ is a piecewise continuous function with improper integrals.

Also see

 * Piecewise Continuous Function with Improper Integrals may not be Bounded