Piecewise Continuous Function does not necessarily have Improper Integrals

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

Let $f$ be a piecewise continuous function with improper integrals:

Then $f$ satisfies:

Definition 4
The converse is not true.

Proof
Let $f$ be a piecewise continuous function with improper integrals.

We need to prove that $f$ satisfies the requirements of Definition 4.

We observe that Definition 4 is subsumed in the definition of a piecewise continuous function with improper integrals.

Also, this part is connected to the rest of Definition 3 by a logical AND.

Therefore, a function satisfying Definition 3 satisfies Definition 4 as well.

Accordingly, Definition 3 implies Definition 4.

Next, we need to prove that Definition 4 is not necessarily piecewise continuous function with improper integrals.

To do this we need a function that satisfies Definition 4 but is not a piecewise continuous function with improper integrals.

We seek a function whose integral $\displaystyle \int_{x_{i - 1}^+}^{x_i^-} f \left({x}\right) \rd x$ does not exist for some $i \in \left\{{1, 2, \ldots, n}\right\}$.

Consider the function:


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

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

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

Therefore, $f$ satisfies the requirements of Definition 4 for the subdivision $\left\{{a, b}\right\}$ of $\left[{a \,.\,.\, b}\right]$.

We now consider $(2)$ of piecewise continuous function with improper integrals, which, in our case, requires that the improper integral $\displaystyle \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ exists.

Let $c$ be a point in $\left({a \,.\,.\, b}\right)$.

From the definition of improper integral, the existence of $\displaystyle \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ requires that $\displaystyle \lim_{\gamma \mathop \to a^+} \int_\gamma^c \dfrac 1 {x - a} \rd x$ exists.

We have

The last right-hand side approaches $\infty$ as $\gamma$ approaches $a$ from above.

So $\displaystyle \int_{a^+}^{c} \dfrac 1 {x - a} \rd x$ does not exist.

Therefore, $\displaystyle \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ does not exist either.

Accordingly, requirement $(2)$ of piecewise continuous function with improper integrals is not satisfied.