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:

Then it is not necessarily the case that $f$ is a piecewise continuous function with improper integrals:

Proof
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, $f$ is a piecewise continuous function for the (finite) subdivision $\left\{{a, b}\right\}$ of $\left[{a \,.\,.\, b}\right]$.

We now consider whether 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 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, $f$ is not a piecewise continuous function with improper integrals.