Equivalence of Definitions of Removable Discontinuity of Real Function
Theorem
Let $A \subseteq \R$ be a subset of the real numbers.
Let $f: A \to \R$ be a real function.
Let $f$ be discontinuous at $a \in A$.
The following definitions of the concept of removable discontinuity are equivalent:
Definition 1
The point $c$ is a removable discontinuity of $f$ if and only if the limit $\ds \lim_{x \mathop \to c} \map f x$ exists.
Definition 2
The point $c$ is a removable discontinuity of $f$ if and only if there exists $b \in \R$ such that the function $f_b$ defined by:
- $\map {f_b} x = \begin {cases} \map f x &: x \ne c \\ b &: x = c \end {cases}$
is continuous at $c$.
Proof
Lemma
Let $A \subseteq \R$ be a subset of the real numbers.
Let $f, g: A \to \R$ be real functions.
Let $a \in A$.
Suppose $\map f x = \map g x$ for every $x \ne a$.
Suppose the limit $\ds \lim_{x \mathop \to a} \map f x$ exists.
Then the limit $\ds \lim_{x \mathop \to a} \map g x$ exists and is equal to $\ds \lim_{x \mathop \to a} \map f x$.
$\Box$
Let $A \subseteq \R$ be a subset of the real numbers.
Let $f: A \to \R$ be a real function.
Let $f$ be discontinuous at $a \in A$.
For any $b \in \R$, define the function $f_b$ by:
- $\map {f_b} x = \begin {cases} \map f x &: x \ne a \\ b &: x = a \end {cases}$
Then $\map {f_b} x = \map f x$ for every $x \ne a$.
Definition 1 implies Definition 2
Suppose the limit $\ds \lim_{x \mathop \to a} \map f x$ exists.
Let $\ds \lim_{x \mathop \to a} \map f x = b$.
Then:
| \(\ds \map {f_b} a\) | \(=\) | \(\ds b\) | Definition of $f_b$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to a} \map f x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to a} \map {f_b} x\) | Lemma |
By definition, $f_b$ is continuous at $a$.
$\Box$
Definition 2 implies Definition 1
Suppose there exists $b \in \R$ such that $f_b$ is continuous at $a$.
- $\ds \lim_{x \mathop \to a} \map {f_b} x = \map {f_b} a = b $
By the lemma:
- $\ds \lim_{x \mathop \to a} \map f x = \lim_{x \mathop \to a} \map {f_b} x = b$
It follows that the limit $\ds \lim_{x \mathop \to a} \map f x$ exists.
$\blacksquare$