Piecewise Continuous Function with One-Sided Limits is Bounded

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Theorem

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

Let $f$ be a piecewise continuous function with one-sided limits.


Then $f$ is a bounded piecewise continuous function.


Proof

Let $f$ be a piecewise continuous function with one-sided limits.

By definition, there exists a finite subdivision $\left\{ {x_0, x_1, \ldots, x_n}\right\}$ of $\left[{a \,.\,.\, b}\right]$, where $x_0 = a$ and $x_n = b$, such that $f$ is continuous on $\left({x_{i − 1} \,.\,.\, x_i}\right)$ for every $i \in \left\{{1, 2, \ldots, n}\right\}$.


For every $i \in \left\{ {1, 2, \ldots, n}\right\}$, we define a function $f_i$ with domain $\left[{x_{i − 1} \,.\,.\, x_i}\right]$, as follows:

$f_i \left({x}\right) := \begin{cases} \displaystyle \lim_{x \mathop \to x_{i − 1}^+} f \left({x}\right) & : x = x_{i - 1} \\ f \left({x}\right) & : x \in \left({x_{i - 1} \,.\,.\, x_i}\right) \\ \displaystyle \lim_{x \mathop \to x_i^-} f \left({x}\right) & : x = x_i \end{cases}$

The one-sided limits in this definition exist because $f$ is piecewise continuous with one-sided limits.


We have that $f$ is continuous on $\left({x_{i − 1} \,.\,.\, x_i}\right)$.

We also have that $f_i = f$ on $\left({x_{i−1} \,.\,.\, x_i}\right)$.

Therefore $f_i$ is also continuous on $\left({x_{i − 1} \,.\,.\, x_i}\right)$.


By definition, $f_i$ is right-continuous at $x_{i − 1}$ and left-continuous at $x_i$.

Therefore, $f_i$ is continuous throughout its domain $\left[{x_{i−1} \,.\,.\, x_i}\right]$.


By Continuous Function on Compact Subspace of Euclidean Space is Bounded and Closed Real Interval is Compact, $f_i$ is bounded.

Therefore, $f_i$ is bounded on $\left[{x_{i−1} \,.\,.\, x_i}\right]$.

We have that $\left({x_{i − 1} \,.\,.\, x_i}\right)$ constitutes a subset of $\left[{x_{i−1} \,.\,.\, x_i}\right]$.

Thus $f_i$ is also bounded on $\left({x_{i − 1} \,.\,.\, x_i}\right)$.


As $f_i = f$ on $\left({x_{i − 1} \,.\,.\, x_i}\right)$, $f$ is also bounded on $\left({x_{i − 1} \,.\,.\, x_i}\right)$.


We have that:

$f$ is bounded on the intervals $\left({x_{i−1} \,.\,.\, x_i}\right)$

and

the set of these intervals is finite.

Hence the set of bounds of $f$ on these intervals is itself bounded.

The bound of this set of bounds serves as a bound for $f$ on each of the intervals $\left({x_{i − 1} \,.\,.\, x_i}\right)$.

Therefore, this bound is a bound for $f$ on the union of these intervals.


That is, $f$ is bounded on the union of $\left({x_{i − 1} \,.\,.\, x_i}\right)$ for all $i \in \left\{{1, 2, \ldots, n}\right\}$.


The only points left to consider are the points in the set $\left\{{x_0, x_1, \ldots, x_n}\right\}$.

Since this set is finite, the maximum $\max \left({\left\vert{f \left({x_0}\right)}\right\vert, \left\vert{f \left({x_1}\right)}\right\vert, \ldots, \left\vert{f \left({x_n}\right)}\right\vert}\right)$ is finite and serves as a bound for $f$ on $\left\{{x_0, x_1, \ldots, x_n}\right\}$.


Thus $f$ is bounded on the whole of its domain $\left[{a \,.\,.\, b}\right]$.

Hence $f$ is a bounded piecewise continuous function.

$\blacksquare$


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