--- title: "Classification of discontinuities" chunk: 3/4 source: "https://en.wikipedia.org/wiki/Classification_of_discontinuities" category: "reference" tags: "science, encyclopedia" date_saved: "2026-05-05T09:08:17.602531+00:00" instance: "kb-cron" --- 1 C ( x ) = { 1 x ∈ C 0 x ∈ [ 0 , 1 ] ∖ C . {\displaystyle \mathbf {1} _{\mathcal {C}}(x)={\begin{cases}1&x\in {\mathcal {C}}\\0&x\in [0,1]\setminus {\mathcal {C}}.\end{cases}}} One way to construct the Cantor set C {\displaystyle {\mathcal {C}}} is given by C := ⋂ n = 0 ∞ C n {\textstyle {\mathcal {C}}:=\bigcap _{n=0}^{\infty }C_{n}} where the sets C n {\displaystyle C_{n}} are obtained by recurrence according to C n = C n − 1 3 ∪ ( 2 3 + C n − 1 3 ) for n ≥ 1 , and C 0 = [ 0 , 1 ] . {\displaystyle C_{n}={\frac {C_{n-1}}{3}}\cup \left({\frac {2}{3}}+{\frac {C_{n-1}}{3}}\right){\text{ for }}n\geq 1,{\text{ and }}C_{0}=[0,1].} In view of the discontinuities of the function 1 C ( x ) , {\displaystyle \mathbf {1} _{\mathcal {C}}(x),} let's assume a point x 0 ∉ C . {\displaystyle x_{0}\not \in {\mathcal {C}}.} Therefore there exists a set C n , {\displaystyle C_{n},} used in the formulation of C {\displaystyle {\mathcal {C}}} , which does not contain x 0 . {\displaystyle x_{0}.} That is, x 0 {\displaystyle x_{0}} belongs to one of the open intervals which were removed in the construction of C n . {\displaystyle C_{n}.} This way, x 0 {\displaystyle x_{0}} has a neighbourhood with no points of C . {\displaystyle {\mathcal {C}}.} (In another way, the same conclusion follows taking into account that C {\displaystyle {\mathcal {C}}} is a closed set and so its complementary with respect to [ 0 , 1 ] {\displaystyle [0,1]} is open). Therefore 1 C {\displaystyle \mathbf {1} _{\mathcal {C}}} only assumes the value zero in some neighbourhood of x 0 . {\displaystyle x_{0}.} Hence 1 C {\displaystyle \mathbf {1} _{\mathcal {C}}} is continuous at x 0 . {\displaystyle x_{0}.} This means that the set D {\displaystyle D} of all discontinuities of 1 C {\displaystyle \mathbf {1} _{\mathcal {C}}} on the interval [ 0 , 1 ] {\displaystyle [0,1]} is a subset of C . {\displaystyle {\mathcal {C}}.} Since C {\displaystyle {\mathcal {C}}} is an uncountable set with null Lebesgue measure, also D {\displaystyle D} is a null Lebesgue measure set and so in the regard of Lebesgue-Vitali theorem 1 C {\displaystyle \mathbf {1} _{\mathcal {C}}} is a Riemann integrable function. More precisely one has D = C . {\displaystyle D={\mathcal {C}}.} In fact, since C {\displaystyle {\mathcal {C}}} is a nonwhere dense set, if x 0 ∈ C {\displaystyle x_{0}\in {\mathcal {C}}} then no neighbourhood ( x 0 − ε , x 0 + ε ) {\displaystyle \left(x_{0}-\varepsilon ,x_{0}+\varepsilon \right)} of x 0 , {\displaystyle x_{0},} can be contained in C . {\displaystyle {\mathcal {C}}.} This way, any neighbourhood of x 0 ∈ C {\displaystyle x_{0}\in {\mathcal {C}}} contains points of C {\displaystyle {\mathcal {C}}} and points which are not of C . {\displaystyle {\mathcal {C}}.} In terms of the function 1 C {\displaystyle \mathbf {1} _{\mathcal {C}}} this means that both lim x → x 0 − 1 C ( x ) {\textstyle \lim _{x\to x_{0}^{-}}\mathbf {1} _{\mathcal {C}}(x)} and lim x → x 0 + 1 C ( x ) {\textstyle \lim _{x\to x_{0}^{+}}1_{\mathcal {C}}(x)} do not exist. That is, D = E 1 , {\displaystyle D=E_{1},} where by E 1 , {\displaystyle E_{1},} as before, we denote the set of all essential discontinuities of first kind of the function 1 C . {\displaystyle \mathbf {1} _{\mathcal {C}}.} Clearly ∫ 0 1 1 C ( x ) d x = 0. {\textstyle \int _{0}^{1}\mathbf {1} _{\mathcal {C}}(x)dx=0.} == Discontinuities of derivatives == Let I ⊆ R {\displaystyle I\subseteq \mathbb {R} } an open interval, let F : I → R {\displaystyle F:I\to \mathbb {R} } be differentiable on I , {\displaystyle I,} and let f : I → R {\displaystyle f:I\to \mathbb {R} } be the derivative of F . {\displaystyle F.} That is, F ′ ( x ) = f ( x ) {\displaystyle F'(x)=f(x)} for every x ∈ I {\displaystyle x\in I} . According to Darboux's theorem, the derivative function f : I → R {\displaystyle f:I\to \mathbb {R} } satisfies the intermediate value property. The function f {\displaystyle f} can, of course, be continuous on the interval I , {\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 f {\displaystyle f} to be continuous and the intermediate value property does not imply f {\displaystyle f} is continuous on I . {\displaystyle I.} Darboux's theorem does, however, have an immediate consequence on the type of discontinuities that f {\displaystyle f} can have. In fact, if x 0 ∈ I {\displaystyle x_{0}\in I} is a point of discontinuity of f {\displaystyle f} , then necessarily x 0 {\displaystyle x_{0}} is an essential discontinuity of f {\displaystyle f} . This means in particular that the following two situations cannot occur: Furthermore, two other situations have to be excluded (see John Klippert): Observe that whenever one of the conditions (i), (ii), (iii), or (iv) is fulfilled for some x 0 ∈ I {\displaystyle x_{0}\in I} one can conclude that f {\displaystyle f} fails to possess an antiderivative, F {\displaystyle F} , on the interval I {\displaystyle I} . On the other hand, a new type of discontinuity with respect to any function f : I → R {\displaystyle f:I\to \mathbb {R} } can be introduced: an essential discontinuity, x 0 ∈ I {\displaystyle x_{0}\in I} , of the function f {\displaystyle f} , is said to be a fundamental essential discontinuity of f {\displaystyle f} if lim x → x 0 − f ( x ) ≠ ± ∞ {\displaystyle \lim _{x\to x_{0}^{-}}f(x)\neq \pm \infty } and lim x → x 0 + f ( x ) ≠ ± ∞ . {\displaystyle \lim _{x\to x_{0}^{+}}f(x)\neq \pm \infty .}