Piecewise Continuous Function does not necessarily have Improper Integrals

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

Then: if $f$ satisfies Piecewise Continuous Function/Definition 3, $f$ satisfies Piecewise Continuous Function/Definition 4.

The converse is not true.

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

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

We observe that Definition 4 is part of Definition 3, and that 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.

To prove that Definition 4 does not imply Definition 3, we need to find a function that satisfies Definition 4 but not Definition 3.

We shall do this by seeking a function whose integral $\displaystyle \int_{x_{i-1}+}^{x_i-} f \left(x\right) \ \mathrm d x$ does not exist for some $i \in \left\{{1, \ldots, n}\right\}$.

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 Definition 3, which, in our case, requires that the improper integral $\displaystyle \int_{a+}^{b-} \dfrac 1 {x-a} \ \mathrm d x$ exist.

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

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

We have

The last right-hand side approaches $\infty$ as $\gamma$ approaches $a$ from above, so $\displaystyle \int_{a+}^{c} \dfrac 1 {x-a} \ \mathrm d x$ does not exist.

Therefore, $\displaystyle \int_{a+}^{b-} \dfrac 1 {x-a} \ \mathrm d x$ does not exist either, and so requirement (2) of Definition 3 is not satisfied.