# 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 $a$ is a **removable discontinuity** of $f$ if and only if the limit $\displaystyle \lim_{x \mathop \to a} \map f x$ exists.

### Definition 2

The point $a$ is a **removable discontinuity** of $f$ if and only if there exists $b \in \R$ such that the function $f_b$ defined by:

- $\map {f_b} x = \begin {cases} \map f x &: x \ne a \\ b &: x = a \end {cases}$

is continuous at $a$.

## 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 $f_b$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to a} \map f x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to a} \map {f_b} x\) | By Lemma |

By definition of continuity, $f_b$ 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 $

Then 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.

$\blacksquare$