Counter example regarding bounded functions.

Question

I feel that I completely missed the point in this question, as I didn't even use / don't know how to use the fact that $a,b$ are removable discontinuities.. Does that mean I need to construct a function $g$ that will cover the scenario regarding the discontinuity points?

Counter example:

Consider the function $f: R\ ➜ R$ , $f\left(x\right)\ =\ \frac{1}{x}$

Let $(a,b) = (o,1)$. Note that $f$ is continuous on $(0,1)$.

$\lim _{x\to 0^+}\left(\frac{1}{x}\right)=∞$

$\lim _{x\to 1^-}\left(\frac{1}{x}\right)=1$

Since the one sided limits tends to $∞$, $f$ is not bounded on $(0,1)$.

Answer 1

Definition: A function $f$ is said to have a removable discontinuity at $a \in \mathbb{R}$, if $\lim_\limits{x \to a} f(x)$ exists finitely and $\lim_\limits{x \to a} f(x)\neq f(a)$.

By this definition, see $\lim_\limits{x \to 0}\frac{1}{x}$ is not finite. So it is not a counterexample.

Letting $\lim_\limits{x \to a}f(x)=p$ and $\lim_\limits{x \to b}f(x)=q$, one can construct a function $g:[a,b] \to \mathbb{R}$ such that $$g(x):= \begin{cases} f(x) & x \in (a,b) \\ p & x=a \\ q & x=b \\ \end{cases}$$ Now $g$ is a bounded function in $[a,b]$.
(this follows from the Bolzano-Weierstrass theorem which in turn follows from lub-axiom)

$g$ is bounded $\implies$ $\exists \ M \in \mathbb{R}$ such that $g(x) \leq M$ for all $x \in [a,b]$ $\implies f(x) \leq M$ for all $x \in (a,b)$, i.e $f$ is bounded in (a,b).

Answer 2

Let $c=\frac{a+b}2$. By definition of "removable discontinuity", there exists a function $g\colon[a,b)\to \Bbb R$ such that $g|_{(a,b)}=f$ and $g$ is continiuous at $a$. it follows that $g$ is continuous, hence is continuous on the interval $[a,c]$, by compactness is bounded there, and $f$ is also bounded on $(a,c]$.

Similarly, the removable discontinuity at $b$ shows that $f$ is bounded on $[c,b)$, hence $f$ is bounded on all of $(a,b)$.