# Classification of discontinuities

Continuous functions are of utmost importance in mathematics, functions and applications. However, not all functions are continuous. If a function is not continuous at a point in its domain, one says that it has a discontinuity there. The set of all points of discontinuity of a function may be a discrete set, a dense set, or even the entire domain of the function. This article describes the classification of discontinuities in the simplest case of functions of a single real variable taking real values.

The oscillation of a function at a point quantifies these discontinuities as follows:

• in a removable discontinuity, the distance that the value of the function is off by is the oscillation;
• in a jump discontinuity, the size of the jump is the oscillation (assuming that the value at the point lies between these limits of the two sides);
• in an essential discontinuity, oscillation measures the failure of a limit to exist; the limit is constant.

A special case is if the function diverges to infinity or minus infinity, in which case the oscillation is not defined (in the extended real numbers, this is a removable discontinuity).

## Classification

For each of the following, consider a real valued function f of a real variable x, defined in a neighborhood of the point x0 at which f is discontinuous.

### Removable discontinuity

Consider the piecewise function

$f(x)={\begin{cases}x^{2}&{\mbox{ for }}x<1\\0&{\mbox{ for }}x=1\\2-x&{\mbox{ for }}x>1\end{cases}}$

The point x0 = 1 is a removable discontinuity. For this kind of discontinuity:

The one-sided limit from the negative direction:

$L^{-}=\lim _{x\to x_{0}^{-}}f(x)$

and the one-sided limit from the positive direction:

$L^{+}=\lim _{x\to x_{0}^{+}}f(x)$

at x0 both exist, are finite, and are equal to L = L = L+. In other words, since the two one-sided limits exist and are equal, the limit L of f(x) as x approaches x0 exists and is equal to this same value. If the actual value of f(x0) is not equal to L, then x0 is called a removable discontinuity. This discontinuity can be removed to make f continuous at x0, or more precisely, the function

$g(x)={\begin{cases}f(x)&x\neq x_{0}\\L&x=x_{0}\end{cases}}$

is continuous at x = x0.

The term removable discontinuity is sometimes broadened to include a removable singularity, in which the limits in both directions exist and are equal, while the function is undefined at the point x0.[a] This use is an abuse of terminology because continuity and discontinuity of a function are concepts defined only for points in the function's domain.

### Jump discontinuity

Consider the function

$f(x)={\begin{cases}x^{2}&{\mbox{ for }}x<1\\0&{\mbox{ for }}x=1\\2-(x-1)^{2}&{\mbox{ for }}x>1\end{cases}}$

Then, the point x0 = 1 is a jump discontinuity.

In this case, a single limit does not exist because the one-sided limits, L and L+, exist and are finite, but are not equal: since, LL+, the limit L does not exist. Then, x0 is called a jump discontinuity, step discontinuity, or discontinuity of the first kind. For this type of discontinuity, the function f may have any value at x0.

### Essential discontinuity

For an essential discontinuity, at least one of the two one-sided limits doesn't exist. Consider the function

$f(x)={\begin{cases}\sin {\frac {5}{x-1}}&{\text{ for }}x<1\\0&{\text{ for }}x=1\\{\frac {1}{x-1}}&{\text{ for }}x>1.\end{cases}}$

Then, the point $x_{0}=1$  is an essential discontinuity.

In this example, both $L^{-}$  and $L^{+}$  don't exist, thus satisfying the condition of essential discontinuity. So x0 is an essential discontinuity, infinite discontinuity, or discontinuity of the second kind. (This is distinct from an essential singularity, which is often used when studying functions of complex variables.)

## The set of discontinuities of a function

Suppose f is a function defined on an interval $I\subset \mathbb {R}$ . We will denote by $D$  the set of all discontinuities on the interval $I$  of the function f . The two following properties of the set $D$  are relevant in the literature.

• The set of $D$  is an Fσ set. The set of points at which a function is continuous is always a Gδ set.
• If on the interval $I$ , f is monotone then $D$  is at most countable. This is Froda's theorem.
• When $I=[a,b]$  and f is a bounded function it is well-known the importance of the set $D$  in the regard of the Riemann integrability of f. In fact, Lebesgue-Vitali theorem states that f is Riemann integrable on $I=[a,b]$  if and only if $D$  has null Lebesgue's measure.

For Lebesgue-Vitali seems that all type of discontinuities have the same weight on the obstruction that a bounded function f be Riemann integrable on $[a,b]$ . However this not the case. In fact, certain discontinuities have absolutly no role on the Riemann integrability of the function. In order to clarify this question , it is worthwhile to divide the set $D$  into the following three sets corresponding to the type of discontinuities initially considered.

For this purpose by $R$  we will mean the set of all $x_{0}\in I$  such that f has a removable discontinuity at $x_{0}$ . Analogously by $J$  we denote the set constituted by all all $x_{0}\in I$  such that f has a jump discontinuity at $x_{0}$ . The set of all $x_{0}\in I$  such that f has an essential discontinuity at $x_{0}$  will be denoted by $E$ . Of course the set $D$  of all discontinuities on the interval $I$  of the function f is such that $D=R\cup J\cup E$ .

The set $E$  of all essential discontinuities can be subdivided into the following two sets:

$E_{1}=\{x_{0}\in E:\lim _{x\to x_{0}^{-}}f(x)$  and $\lim _{x\to x_{0}^{+}}f(x)$  don't exist$\}$  and $E_{2}=\{x_{0}\in E:$  either $\lim _{x\to x_{0}^{-}}f(x)$  or $\lim _{x\to x_{0}^{+}}f(x)$  exist$\}$ .

Of course $E=E_{1}\cup E_{2}$ . Whenever $x_{0}\in E_{1}$  we will say that $x_{0}$  is an essential discontinuity of first kind. For any $x_{0}\in E_{2}$  we will call it an essential discontinuity of second kind.

One has the following important property (see ):

• The set $R\cup J\cup E_{2}$  is countable.

Therefore the Lebesgue-Vitali theorem can be rephrased as follows:

• A bounded function, f , is Riemann integrable on $[a,b]$  if and only if the correspondent set $E_{1}$  of all essential discontinuities of first kind of f has null Lebesgue's measure.

Thomae's function is discontinuous at every non-zero rational point, but continuous at every irrational point. One easily sees that those discontinuities are all essential of the first kind. By the first paragraph, there does not exist a function that is continuous at every rational point, but discontinuous at every irrational point.

The indicator function of the rationals, also known as the Dirichlet function, is discontinuous everywhere. These discontinuities are all essential of the first kind too.

## Example

Consider now the Cantor set ${\mathcal {C}}\subset [0,1]$  and its indicator (or characteristic) function

$\mathbf {1} _{\mathcal {C}}(x)={\begin{cases}1&x\in {\mathcal {C}}\\0&x\notin [0,1]\setminus {\mathcal {C}}.\end{cases}}$

In view of the discontinuities of this function, let's assume a point $x_{0}\not \in {\mathcal {C}}$ . Therefore there exists a set $C_{n}$ , used in the formulation of ${\mathcal {C}}$ , which does not contain $x_{0}$ . That is, $x_{0}$  belongs to one of the open intervals which were removed in the construction of $C_{n}$ . This way, $x_{0}$  has a neighbourhood with no points of ${\mathcal {C}}$ , where by consequence $\mathbf {1} _{\mathcal {C}}(x)$  only assumes the value zero. Hence $\mathbf {1} _{\mathcal {C}}$  is continuous in $x_{0}$ . This means that the set $D$  of all discontinuities of $\mathbf {1} _{\mathcal {C}}$  on the interval $[0,1]$  is a subset of ${\mathcal {C}}$ . Since ${\mathcal {C}}$  is a noncountable set with null Lebesgue measure, also $D$  is a null Lebesgue measure set and so in the regard of Lebesgue-Vitali theorem $\mathbf {1} _{\mathcal {C}}$  is a Riemann integrable function.

But more precisely one has $D={\mathcal {C}}$ . In fact, if $x_{0}\in {\mathcal {C}}$ , no neighbourhoof ${\bigl (}x_{0}-\epsilon ,x_{0}+\epsilon {\bigr )}$  of $x_{0}$  can be contained in ${\mathcal {C}}$ . Otherwise we should have, for every $n$ , ${\bigl (}x_{0}-\epsilon ,x_{0}+\epsilon {\bigr )}\subset C_{n}$ , which is absurd. since eachone of these sets is composed by interval with length $3^{-n}$ , which doesn't allow that inclusion for values of $n$  sufficiently large in manner that $3^{-n}<\epsilon$ . This way any neighbourhood of $x_{0}\in {\mathcal {C}}$  contains points of ${\mathcal {C}}$  and points which are not of ${\mathcal {C}}$ . In terms of the function $\mathbf {1} _{\mathcal {C}}$  this means that both $\lim _{x\to x_{0}^{-}}\mathbf {1} _{\mathcal {C}}(x)$  and $\lim _{x\to x_{0}^{+}}1_{\mathcal {C}}(x)$  don´t exist. That is, $D=E_{1}$ , where by $E_{1}$ , as before, we denote the set of all essential discontinuities of first kind of the function $\mathbf {1} _{\mathcal {C}}$ . Clearly $\int _{0}^{1}\mathbf {1} _{\mathcal {C}}(x)dx=0.$