Classification of discontinuities

Counting discontinuities of a function

Supposing that ${\displaystyle f}$ is a function defined on an interval ${\displaystyle I\subseteq \mathbb{R},}$ we will denote by ${\displaystyle D}$ the set of all discontinuities of ${\displaystyle f}$ on ${\displaystyle I.}$ By ${\displaystyle R}$ we will mean the set of all ${\displaystyle x_0\in I}$ such that ${\displaystyle f}$ has a removable discontinuity at ${\displaystyle x_0.}$ Analogously by ${\displaystyle J}$ we denote the set constituted by all ${\displaystyle x_0\in I}$ such that ${\displaystyle f}$ has a jump discontinuity at ${\displaystyle x_0.}$ The set of all ${\displaystyle x_0\in I}$ such that ${\displaystyle f}$ has an essential discontinuity at ${\displaystyle x_0}$ will be denoted by ${\displaystyle E.}$ Of course then ${\displaystyle D=R\cup J\cup E.}$

The two following properties of the set ${\displaystyle D}$ are relevant in the literature.

• The set ${\displaystyle D}$ is an ${\displaystyle F_{\sigma }}$ set. The set of points at which a function is continuous is always a ${\displaystyle G_{\delta }}$ set (see).
• If on the interval ${\displaystyle I,}$ ${\displaystyle f}$ is monotone then ${\displaystyle D}$ is at most countable and ${\displaystyle D=J.}$ This is Froda's theorem.

Tom Apostol follows partially the classification above by considering only removable and jump discontinuities. His objective is to study the discontinuities of monotone functions, mainly to prove Froda’s theorem. With the same purpose, Walter Rudin and Karl R. Stromberg study also removable and jump discontinuities by using different terminologies. However, furtherly, both authors state that ${\displaystyle R\cup J}$ is always a countable set (see).

The term essential discontinuity has evidence of use in mathematical context as early as 1889. However, the earliest use of the term alongside a mathematical definition seems to have been given in the work by John Klippert. Therein, Klippert also classified essential discontinuities themselves by subdividing the set ${\displaystyle E}$ into the three following sets:

$${\displaystyle \begin{array}{cc}{E}_1& =\{x_0\in I:\underset{x\to x_0^{-}}{lim}f(x)\text{ and }\underset{x\to x_0^{+}}{lim}f(x)\text{ do not exist in }\mathbb{R}\},\\ E_2& =\{x_0\in I:\underset{x\to x_0^{-}}{lim}f(x)\text{ exists in }\mathbb{R}\text{ and }\underset{x\to x_0^{+}}{lim}f(x)\text{ does not exist in }\mathbb{R}\},\\ E_3& =\{x_0\in I:\underset{x\to x_0^{-}}{lim}f(x)\text{ does not exist in }\mathbb{R}\text{ and }\underset{x\to x_0^{+}}{lim}f(x)\text{ exists in }\mathbb{R}\}.\end{array}}$$

Of course ${\displaystyle E=E_1\cup E_2\cup E_3.}$ Whenever ${\displaystyle x_0\in E_1,}$ ${\displaystyle x_0}$ is called an essential discontinuity of first kind. Any ${\displaystyle x_0\in E_2\cup E_3}$ is said an essential discontinuity of second kind. Hence he enlarges the set ${\displaystyle R\cup J}$ without losing its characteristic of being countable, by stating the following:

• The set ${\displaystyle R\cup J\cup E_2\cup E_3}$ is countable.

Rewriting Lebesgue's theorem

When ${\displaystyle I=[a,b]}$ and ${\displaystyle f}$ is a bounded function, it is well-known of the importance of the set ${\displaystyle D}$ in the regard of the Riemann integrability of ${\displaystyle f.}$ In fact, Lebesgue's theorem (also named Lebesgue-Vitali theorem) states that ${\displaystyle f}$ is Riemann integrable on ${\displaystyle I=[a,b]}$ if and only if ${\displaystyle D}$ is a set with Lebesgue's measure zero.

In this theorem seems that all type of discontinuities have the same weight on the obstruction that a bounded function ${\displaystyle f}$ be Riemann integrable on ${\displaystyle [a,b].}$ Since countable sets are sets of Lebesgue's measure zero and a countable union of sets with Lebesgue's measure zero is still a set of Lebesgue's measure zero, we are seeing now that this is not the case. In fact, the discontinuities in the set ${\displaystyle R\cup J\cup E_2\cup E_3}$ are absolutely neutral in the regard of the Riemann integrability of ${\displaystyle f.}$ The main discontinuities for that purpose are the essential discontinuities of first kind and consequently the Lebesgue-Vitali theorem can be rewritten as follows:

• A bounded function, ${\displaystyle f,}$ is Riemann integrable on ${\displaystyle [a,b]}$ if and only if the correspondent set ${\displaystyle E_1}$ of all essential discontinuities of first kind of ${\displaystyle f}$ has Lebesgue's measure zero.

The case where ${\displaystyle E_1=\varnothing }$ correspond to the following well-known classical complementary situations of Riemann integrability of a bounded function ${\displaystyle f:[a,b]\to \mathbb{R}}$:

• If ${\displaystyle f}$ has right-hand limit at each point of ${\displaystyle [a,b[}$ then ${\displaystyle f}$ is Riemann integrable on ${\displaystyle [a,b]}$ (see)
• If ${\displaystyle f}$ has left-hand limit at each point of ${\displaystyle ]a,b]}$ then ${\displaystyle f}$ is Riemann integrable on ${\displaystyle [a,b].}$
• If ${\displaystyle f}$ is a regulated function on ${\displaystyle [a,b]}$ then ${\displaystyle f}$ is Riemann integrable on ${\displaystyle [a,b].}$

Examples

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 removable. 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.

Consider now the ternary Cantor set ${\displaystyle \mathcal{C}\subset [0,1]}$ and its indicator (or characteristic) function $${\displaystyle {\mathbf{1}}_{\mathcal{C}}(x)=\{\begin{array}{ll}1& x\in \mathcal{C}\\ 0& x\in [0,1]\setminus \mathcal{C}.\end{array}}$$ One way to construct the Cantor set ${\displaystyle \mathcal{C}}$ is given by $\mathcal{C}:=\bigcap _{n=0}^{\mathrm{\infty }}{C}_n$ where the sets ${\displaystyle C_n}$ are obtained by recurrence according to $${\displaystyle C_n=\frac{C_{n-1}}{3}\cup \left(\frac{2}{3}+\frac{C_{n-1}}{3}\right)\text{ for }n\ge 1,\text{ and }{C}_0=[0,1].}$$

In view of the discontinuities of the function ${\displaystyle {\mathbf{1}}_{\mathcal{C}}(x),}$ let's assume a point ${\displaystyle x_0\notin \mathcal{C}.}$

Therefore there exists a set ${\displaystyle C_n,}$ used in the formulation of ${\displaystyle \mathcal{C}}$, which does not contain ${\displaystyle x_0.}$ That is, ${\displaystyle x_0}$ belongs to one of the open intervals which were removed in the construction of ${\displaystyle C_n.}$ This way, ${\displaystyle x_0}$ has a neighbourhood with no points of ${\displaystyle \mathcal{C}.}$ (In another way, the same conclusion follows taking into account that ${\displaystyle \mathcal{C}}$ is a closed set and so its complementary with respect to ${\displaystyle [0,1]}$ is open). Therefore ${\displaystyle {\mathbf{1}}_{\mathcal{C}}}$ only assumes the value zero in some neighbourhood of ${\displaystyle x_0.}$ Hence ${\displaystyle {\mathbf{1}}_{\mathcal{C}}}$ is continuous at ${\displaystyle x_0.}$

This means that the set ${\displaystyle D}$ of all discontinuities of ${\displaystyle {\mathbf{1}}_{\mathcal{C}}}$ on the interval ${\displaystyle [0,1]}$ is a subset of ${\displaystyle \mathcal{C}.}$ Since ${\displaystyle \mathcal{C}}$ is an uncountable set with null Lebesgue measure, also ${\displaystyle D}$ is a null Lebesgue measure set and so in the regard of Lebesgue-Vitali theorem ${\displaystyle {\mathbf{1}}_{\mathcal{C}}}$ is a Riemann integrable function.

More precisely one has ${\displaystyle D=\mathcal{C}.}$ In fact, since ${\displaystyle \mathcal{C}}$ is a nonwhere dense set, if ${\displaystyle x_0\in \mathcal{C}}$ then no neighbourhood ${\displaystyle \left(x_0-\epsilon ,x_0+\epsilon \right)}$ of ${\displaystyle x_0,}$ can be contained in ${\displaystyle \mathcal{C}.}$ This way, any neighbourhood of ${\displaystyle x_0\in \mathcal{C}}$ contains points of ${\displaystyle \mathcal{C}}$ and points which are not of ${\displaystyle \mathcal{C}.}$ In terms of the function ${\displaystyle {\mathbf{1}}_{\mathcal{C}}}$ this means that both $\underset{x\to x_0^{-}}{lim}{\mathbf{1}}_{\mathcal{C}}(x)$ and $\underset{x\to x_0^{+}}{lim}{1}_{\mathcal{C}}(x)$ do not exist. That is, ${\displaystyle D=E_1,}$ where by ${\displaystyle E_1,}$ as before, we denote the set of all essential discontinuities of first kind of the function ${\displaystyle {\mathbf{1}}_{\mathcal{C}}.}$ Clearly ${\int }_0^1{\mathbf{1}}_{\mathcal{C}}(x)dx=0.$

Discontinuities of derivatives

Let ${\displaystyle I\subseteq \mathbb{R}}$ an open interval, let ${\displaystyle F:I\to \mathbb{R}}$ be differentiable on ${\displaystyle I,}$ and let ${\displaystyle f:I\to \mathbb{R}}$ be the derivative of ${\displaystyle F.}$ That is, ${\displaystyle F^{\prime }(x)=f(x)}$ for every ${\displaystyle x\in I}$. According to Darboux's theorem, the derivative function ${\displaystyle f:I\to \mathbb{R}}$ satisfies the intermediate value property. The function ${\displaystyle f}$ can, of course, be continuous on the interval ${\displaystyle I,}$ in which case Bolzano's theorem also applies. Recall that Bolzano's theorem asserts that every continuous function satisfies the intermediate value property. On the other hand, the converse is false: Darboux's theorem does not assume ${\displaystyle f}$ to be continuous and the intermediate value property does not imply ${\displaystyle f}$ is continuous on ${\displaystyle I.}$

Darboux's theorem does, however, have an immediate consequence on the type of discontinuities that ${\displaystyle f}$ can have. In fact, if ${\displaystyle x_0\in I}$ is a point of discontinuity of ${\displaystyle f}$, then necessarily ${\displaystyle x_0}$ is an essential discontinuity of ${\displaystyle f}$. This means in particular that the following two situations cannot occur:

1. ${\displaystyle x_0}$ is a removable discontinuity of ${\displaystyle f}$.
2. ${\displaystyle x_0}$ is a jump discontinuity of ${\displaystyle f}$.

Furthermore, two other situations have to be excluded (see John Klippert):

1. ${\displaystyle \underset{x\to x_0^{-}}{lim}f(x)=\pm \mathrm{\infty }.}$
2. ${\displaystyle \underset{x\to x_0^{+}}{lim}f(x)=\pm \mathrm{\infty }.}$

Observe that whenever one of the conditions (i), (ii), (iii), or (iv) is fulfilled for some ${\displaystyle x_0\in I}$ one can conclude that ${\displaystyle f}$ fails to possess an antiderivative, ${\displaystyle F}$, on the interval ${\displaystyle I}$.

On the other hand, a new type of discontinuity with respect to any function ${\displaystyle f:I\to \mathbb{R}}$ can be introduced: an essential discontinuity, ${\displaystyle x_0\in I}$, of the function ${\displaystyle f}$, is said to be a fundamental essential discontinuity of ${\displaystyle f}$ if

$${\displaystyle \underset{x\to x_0^{-}}{lim}f(x)\ne \pm \mathrm{\infty }}$$ and $${\displaystyle \underset{x\to x_0^{+}}{lim}f(x)\ne \pm \mathrm{\infty }.}$$

Therefore if ${\displaystyle x_0\in I}$ is a discontinuity of a derivative function ${\displaystyle f:I\to \mathbb{R}}$, then necessarily ${\displaystyle x_0}$ is a fundamental essential discontinuity of ${\displaystyle f}$.

Notice also that when ${\displaystyle I=[a,b]}$ and ${\displaystyle f:I\to \mathbb{R}}$ is a bounded function, as in the assumptions of Lebesgue's theorem, we have for all ${\displaystyle x_0\in (a,b)}$: $${\displaystyle \underset{x\to x_0^{\pm }}{lim}f(x)\ne \pm \mathrm{\infty },}$$ $${\displaystyle \underset{x\to a^{+}}{lim}f(x)\ne \pm \mathrm{\infty },}$$ and $${\displaystyle \underset{x\to b^{-}}{lim}f(x)\ne \pm \mathrm{\infty }.}$$ Therefore any essential discontinuity of ${\displaystyle f}$ is a fundamental one.