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.

The converse is not necessarily true.

Bounded Piecewise Continuous Function has Improper Integrals
Let $f$ be piecewise continuous and bounded on $\left[{a \,.\,.\, b}\right]$.

We need to prove that $f$ satisfies the requirements to be a piecewise continuous function with improper integrals.

The only difference between piecewise continuous and bounded and piecewise continuous function with improper integrals lies in $(2)$ in the two definitions.

So it is sufficient to prove $(2)$ in the definition of a piecewise continuous function with improper integrals.

This requirement states that the improper integral $\displaystyle \int_{x_{i-1}+}^{x_i-} f \left({x}\right) \rd x$ exists for every $i \in \left\{{1, \ldots, n}\right\}$.

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

Therefore, we need to prove 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.

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

This concludes the proof that a bounded piecewise continuous function is a piecewise continuous function with improper integrals.

Piecewise Continuous Function with Improper Integrals is not necessarily Bounded
We need to prove that $f$ being a piecewise continuous function with improper integrals does not imply that $f$ is necessarily also a bounded piecewise continuous function.

We do this by defining a piecewise continuous function with improper integrals which is not bounded.

Consider the function:


 * $f \left({x}\right) = \begin{cases}

0 & x = a \\ \dfrac 1 {\sqrt{x - a} } & x \in \left({a \,.\,.\, b}\right] \end{cases}$

Since $\dfrac 1 {\sqrt{x - a} }$ is continuous on $\left({a \,.\,.\, b}\right)$, $f$ is continuous on $\left({a \,.\,.\, b}\right)$.

Therefore, $f$ satisfies $(1)$ in the requirements of a piecewise continuous function with improper integrals for the subdivision $\left\{{a, b}\right\}$ of $\left[{a \,.\,.\, b}\right]$.

Also,

From this we gather that $\displaystyle \int_{a+}^{b-} f \left({x}\right) \rd x$ exists.

This gives that $f$ is a piecewise continuous function with improper integrals.

However, since $f \left({x}\right)$ approaches $\infty$ as $x$ approaches $a$ from above, $f$ is not bounded.

Therefore, $f$ is not piecewise continuous and bounded.

This completes the proof.