Bounded Piecewise Continuous Function has Improper Integrals

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

Then: if $f$ satisfies Piecewise Continuous Function/Definition 2, $f$ satisfies Piecewise Continuous Function/Definition 3.

The converse is not true.

Definition 2 implies Definition 3
Assume first that $f$ satisfies the requirements of Definition 2.

We need to prove that $f$ satisfies the requirements of Definition 3.

The only difference between Definition 2 and Definition 3 lies in (2) in the two definitions, so it is sufficient to prove (2) in Definition 3, which requires that the improper integrals $\displaystyle \int_{x_{i-1}+}^{x_i-} f \left(x\right) \ \mathrm d x$ exist for every $i \in \left\{{1, \ldots, n}\right\}$.

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

By definition, the improper integral $\displaystyle \int_{x_{i-1}+}^{x_i-} f \left(x\right) \ \mathrm d x$ exists iff $\displaystyle \lim_{\gamma \mathop \to x_{i-1}+} \int_\gamma^c f \left( x \right) \ \mathrm d x$ and $\displaystyle \lim_{\gamma \mathop \to x_i-} \int_c^\gamma f \left( x \right) \ \mathrm d x$ exist.

Therefore, we need to prove that $\displaystyle \lim_{\gamma \mathop \to x_i-} \int_c^\gamma f \left( x \right) \ \mathrm d x$ and $\displaystyle \lim_{\gamma \mathop \to x_{i-1}+} \int_\gamma^c f \left( x \right) \ \mathrm d x$ exist.

By Piecewise Continuous Function/Definition 2 is Riemann Integrable we know that $f$ is integrable on $\left[{a \,.\,.\, b}\right]$ and, therefore, on every closed subinterval of $\left[{a \,.\,.\, b}\right]$.

Accordingly, the definite integrals $\displaystyle \int_c^{x_i} f \left( x \right) \ \mathrm d x$, $\displaystyle \int_c^\gamma f \left( x \right) \ \mathrm d x$, $\displaystyle \int_\gamma^c f \left( x \right) \ \mathrm d x$, and $\displaystyle \int_{x_{i-1}}^c f \left( x \right) \ \mathrm d x$ exist.

We have

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

This shows that $\displaystyle \lim_{\gamma \mathop \to x_i-} \int_c^\gamma f \left( x \right) \ \mathrm d x$ equals $\displaystyle \int_c^{x_i} f \left( x \right) \ \mathrm d x$, and from this we gather that $\displaystyle \lim_{\gamma \mathop \to x_i-} \int_c^\gamma f \left( x \right) \ \mathrm d x$ exists.

Also,

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

This shows that $\displaystyle \lim_{\gamma \mathop \to x_{i-1}+} \int_\gamma^c f \left( x \right) \ \mathrm d x$ equals $\displaystyle \int_{x_{i-1} }^c f \left( x \right) \ \mathrm d x$, and from this we gather that $\displaystyle \lim_{\gamma \mathop \to x_{i-1}+} \int_\gamma^c f \left( x \right) \ \mathrm d x$ exists.

Since $i$ is arbirtrary, this concludes the proof of that $\displaystyle \lim_{\gamma \mathop \to x_i-} \int_c^\gamma f \left( x \right) \ \mathrm d x$ and $\displaystyle \lim_{\gamma \mathop \to x_{i-1}+} \int_\gamma^c f \left( x \right) \ \mathrm d x$ exist for every $i \in \left\{{1, \ldots, n}\right\}$, that $\displaystyle \int_{x_{i-1}+}^{x_i-} f \left(x\right) \ \mathrm d x$ exists for every $i \in \left\{{1, \ldots, n}\right\}$, and that Definition 2 implies Definition 3.

The converse is not true
To prove that Definition 3 does not imply Definition 2, we need to find a function that satisfies Definition 3 but not Definition 2.

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 $\left({a \,.\,.\, b}\right)$ and, therefore, satisfies (1) in the requirements of Definition 3 for the subdivision $\left\{{a, b}\right\}$ of $\left[{a \,.\,.\, b}\right]$.

Also,

From this we gather that $\displaystyle \int_{a+}^{b-} f \left( x \right) \ \mathrm d x$ exists which gives that $f$ satisfies requirement (2) of Definition 3, and this concludes the proof that $f$ satisfies Definition 3.

However, since $f \left({x}\right)$ approaches $\infty$ as $x$ approaches $a$ from above, $f$ is not bounded and does, therefore, not satisfy (2) in Definition 2, and this completes the proof.