15 KiB
| title | chunk | source | category | tags | date_saved | instance |
|---|---|---|---|---|---|---|
| Classification of discontinuities | 3/4 | https://en.wikipedia.org/wiki/Classification_of_discontinuities | reference | science, encyclopedia | 2026-05-05T09:08:17.602531+00:00 | 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 .}