Bounded Piecewise Continuous Function has Improper Integrals

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Theorem

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

Let $f$ be piecewise continuous and bounded on $\closedint a b$.


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


Proof

Let $f$ be piecewise continuous and bounded on $\closedint a b$.

It is sufficient to prove that the improper integral $\displaystyle \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists for every $i \in \set {1, 2, \ldots, n}$.


Let $i \in \set {1, 2, \ldots, n}$.

Let $c$ be a point in $\openint {x_{i − 1} } {x_i}$.


By definition, the improper integral $\displaystyle \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists if and only if:

$\displaystyle \lim_{\gamma \mathop \to {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$

and:

$\displaystyle \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$

both exist.


By Bounded Piecewise Continuous Function is Riemann Integrable we know that $f$ is integrable on $\closedint a b$.

Therefore, $f$ is integrable on every closed subinterval of $\closedint a b$.

Accordingly, the following definite integrals:

$\displaystyle \int_{x_{i - 1} }^c \map f x \rd x$
$\displaystyle \int_{\gamma_1}^c \map f x \rd x$
$\displaystyle \int_c^{\gamma_2} \map f x \rd x$
$\displaystyle \int_c^{x_i} \map f x \rd x$

all exist where $\gamma_1 \in \openint {x_{i − 1} } c$ and $\gamma_2 \in \openint c {x_i}$.

Note that $f$ is bounded on $\closedint a b$ as $f$ is piecewise continuous and bounded.

Therefore, a bound $B$ exists for $f$ on $\closedint a b$.


We have:

\(\displaystyle \size {\int_{\gamma_1}^c \map f x \rd x - \int_{x_{i - 1} }^c \map f x \rd x}\) \(=\) \(\displaystyle \size {\int_{x_{i - 1} }^{\gamma_1} \map f x \rd x}\)
\(\displaystyle \) \(\le\) \(\displaystyle B \size {\gamma_1 - x_{i - 1} }\) Corollary to Upper and Lower Bounds of Integral

which approaches $0$ as $\gamma_1$ approaches $x_{i − 1}$.


This shows that $\displaystyle \lim_{\gamma \mathop \to x_{i - 1}+} \int_\gamma^c \map f x \rd x$ equals $\displaystyle \int_{x_{i - 1} }^c \map f x \rd x$.

From this we gather that $\displaystyle \lim_{\gamma \mathop \to {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$ exists.


Also:

\(\displaystyle \size {\int_c^{x_i} \map f x \rd x - \int_c^{\gamma_2} \map f x \rd x}\) \(=\) \(\displaystyle \size {\int_{\gamma_2}^{x_i} \map f x \rd x}\)
\(\displaystyle \) \(\le\) \(\displaystyle B \size {x_i - \gamma_2}\) Corollary to Upper and Lower Bounds of Integral

which approaches $0$ as $\gamma_2$ approaches $x_i$.


This shows that $\displaystyle \lim_{\gamma \mathop \to x_i-} \int_c^\gamma \map f x \rd x$ equals $\displaystyle \int_c^{x_i} \map f x \rd x$.

From this we gather that $\displaystyle \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$ exists.


Since $i$ is arbitrary, we have shown that $\displaystyle \lim_{\gamma \mathop \to {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$ and $\displaystyle \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$ exist for every $i \in \left\{{1, \ldots, n}\right\}$.

Therefore, $\displaystyle \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists for every $i \in \set {1, 2, \ldots, n}$.

That is, $f$ is a piecewise continuous function with improper integrals.

$\blacksquare$


Also see