Piecewise Continuous Function with Improper Integrals may not be Bounded

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Theorem

Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$.

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


Then $f$ may not be piecewise continuous and bounded on $\closedint a b$.


Proof

Consider the function:

$\map f x = \begin{cases} 0 & : x = a \\ \dfrac 1 {\sqrt{x - a} } & : x \in \hointl a b \end{cases}$

Since $\dfrac 1 {\sqrt{x - a} }$ is continuous on $\openint a b$, $f$ is continuous on $\openint a b$.

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


Also:

\(\displaystyle \int_{a+}^{b-} \map f x \rd x\) \(=\) \(\displaystyle \int_{a+}^{b-} \dfrac 1 {\sqrt{x - a} } \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \bigintlimits {2 \sqrt {x - a} } {a+} {b-}\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{x \mathop \to b-} 2 \sqrt{x - a} - \lim_{x \mathop \to a+} 2 \sqrt{x - a}\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \sqrt {b - a} - 2 \sqrt {a - a}\) as $2 \sqrt{x - a}$ is left-continuous at $b$ and right-continuous at $a$
\(\displaystyle \) \(=\) \(\displaystyle 2 \sqrt{b - a}\)

Hence $\displaystyle \int_{a+}^{b-} \map f x \rd x$ exists.

Thus $f$ is a piecewise continuous function with improper integrals.


However, we have that $\map f x$ approaches $\infty$ as $x$ approaches $a$ from above.

Thus $f$ is not bounded.

Therefore $f$ is not piecewise continuous and bounded.

Hence the result.

$\blacksquare$


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