Derivatives Are Discontinuous At Meager Sets:A Freshman’s Proof

Spyridon Tzimas

Abstract. We prove that derivatives are discontinuous at meager sets fol-

lowing an entirely different path than the one followed by Baire in his original

proof over a century ago, all the while maintaining the level of difficulty of theproof. Its individual pieces have been rearranged and accompanied by minimal

narration in order to reveal the reasoning behind each move and ultimatelyestablish a feel of why the statement is true.

Functions in general can be all kinds of discontinuous, but what about deriva-tives in particular? Can they be discontinuous at all to begin with? The answer isyes, as is demonstrated by this simple example [1], they just tend to not show upin applications. So there is a point in investigating how much discontinuous theycan get. We will do this by finding out which cases of discontinuity deny functionsan antiderivative.

In This Paper. All sets are non-empty unless stated otherwise. For a givenset S, we use S◦ to denote its interior and S− to denote its closure. We denotea subinterval of R by I or J . Since (I◦)− = I− (when I is not a unit set) and(J−)◦ = J◦, for convenience in notation, we denote an open I and its closure by I◦

and I− respectively and a closed J and its interior by J− and J◦ respectively.

We begin our investigation with the most discontinuous of functions; the onesthat are discontinuous everywhere. The most well-known example of these is mostprobably Dirichlet’s function [2] and it does not have an antiderivative. This is dueto Darboux’ Theorem.

Theorem 1 (Darboux’s Theorem). Let f : I◦ → R be a function that isdifferentiable everywhere. Then f ′ has the intermediate value property, that is:

∀x1, x2 ∈ I◦ : ∀y ∈ (min{f ′(x1), f ′(x2)},max{f ′(x1), f ′(x2)}) :: ∃x ∈ (min{x1, x2},max{x1, x2}) : f ′(x) = y

Definition. A function f : I◦ → R is a Darboux function, if and only if it hasthe intermediate value property (i.v.p.).

So for a function to be have an antiderivative, it must have the i.v.p.. Are therefunctions that are discontinuous everywhere and have the i.v.p. at the same time?As it turns out, there are. The construction that demonstrated this is called theConway base 13 function [3]. It bypasses Darboux’ Theorem, but this doesn’t meanthat a function whose graph looks like a white paper turned black by painting dotsall over it can actually support an antiderivative...

1

2 SPYRIDON TZIMAS

Definition. It is dense in I◦, if and only if ∀I ′◦ ⊆ I◦ : ∃x ∈ S : x ∈ I ′◦.A set S ⊆ R is nowhere-dense, if and only if ∀I◦ : ∃I ′◦ ⊆ I◦ : ∀x ∈ S : x /∈ I ′◦.Definition. A function f : I → R is a Conway function, if and only if

∃J− that is not a unit set : ∀y ∈ J− : f−1({y}) is dense in I.

Theorem 2. Let f : I ′′◦ → R be a Conway function. Then f does not have anantiderivative.

Proof. Let F be an antiderivative of f . Then ∀x ∈ I ′′◦ : ∀ε > 0 : ∃δ(ε, x) > 0 :

∀x′ ∈ I ′′◦ ∩ (x− δ(ε, x), x+ δ(ε, x)) \ {x} :

∣∣∣∣F (x′)− F (x)

x′ − x− f(x)

∣∣∣∣ < ε⇔

⇔

∀x′ ∈ I ′′◦ ∩ (x, x+ δ(ε, x)) :: (f(x)− ε)(x′ − x) + F (x) < F (x′) < F (x) + (f(x) + ε)(x′ − x)

∀x′ ∈ I ′′◦ ∩ (x− δ(ε, x), x) :: (f(x) + ε)(x′ − x) + F (x) < F (x′) < F (x) + (f(x)− ε)(x′ − x)

Let {ζ, η} = ∂J−. Then, for ε0 = 1

4 |ζ − η|, there is a sequence (xn) ∈ RN such

that x0 ∈ I ′′◦ ∩ f−1({ζ}), x1 ∈ (x0,min{x0 + δ(ε0, x0), sup I ′′◦}) ∩ f−1({η}), and

xn ∈

(max{xn−1 − δ(ε0, xn−1), 12 (xn−2 + xn−1)}, xn−1) ∩ f−1({ζ}),

if n is even

(xn−1,min{xn−1 + δ(ε0, xn−1), 12 (xn−2 + xn−1)}) ∩ f−1({η}),if n is odd

for all n ∈ N∩ [2,+∞). It can be easily shown that (xn) is a Cauchy sequence, so itconverges to a certain ξ ∈ I ′′◦. Then ξ ∈ I ′′◦∩ (xn− δ(ε0, xn), xn + δ(ε0, xn))\{xn}for all n ∈ N; otherwise, the sequence would fail to converge to ξ, a contradiction.Moreover, ∃n0 ∈ N : ∀n ∈ N∩ (n0,+∞) : xn ∈ I ′′◦ ∩ (ξ− δ(ε0, ξ), ξ+ δ(ε0, ξ)) \ {ξ}.

Let n1, n2 ∈ N ∩ (n0,+∞) and be even and odd respectively. Let xn1< ξ.

Then

(f(xn1)− ε0)(ξ − xn1) + F (xn1) < F (ξ) < F (xn1) + (f(xn1) + ε0)(ξ − xn1)(f(ξ) + ε0)(xn1

− ξ) + F (ξ) < F (xn1) < F (ξ) + (f(ξ)− ε0)(xn1

− ξ)

}⇒

⇒ (f(xn1)− 2ε0 − f(ξ))(ξ − xn1

) < 0 < (f(xn1) + 2ε0 − f(ξ))(ξ − xn1

)⇔⇔ |f(ξ)− f(xn1)| < 2ε0

This means that |f(ξ) − ζ| < 12 |ζ − η|. Likewise if ξ < xn1

. Likewise for xn2. So

|ζ−η| = |ζ−f(ξ)+f(ξ)−η| ≤ |f(ξ)− ζ|+ |f(ξ)−η| < |ζ−η|, a contradiction. �

Still, not all Darboux functions that are discontinuous everywhere are Conwayfunctions. So how do they look like, in general? Discontinuity and the i.v.p. puttogether definitely feel like a recipe for a white-noise graph, so they should not betoo far off from being Conway functions. We will prove this, relying heavily onCantor’s Intersection Theorem.

Theorem 3 (Cantor’s Intersection Theorem). Let (I−n : n ∈ N) be a sequencesuch that I−n+1 ⊆ I−n for all n ∈ N. Then there is a certain I−∞ such that

I−∞ =⋂n∈N

I−n .

If limn→+∞

(max I−n −min I−n ) = 0, then there is a certain ξ ∈ R such that I−∞ = {ξ}.

DERIVATIVES ARE DISCONTINUOUS AT MEAGER SETS 3

Theorem 4. Let f : I◦ → R be a Darboux function that is discontinuouseverywhere. Then there is a certain I ′′◦ ⊆ I◦ such that f |I ′′◦ is a Conway function.

Proof. Let f−1({y}) be nowhere-dense in I◦ for all y ∈ R. This means that

∀y ∈ R : ∀I ′◦ ⊆ I◦ : ∃I ′′◦ ⊆ I ′◦ : ∀x ∈ I ′′◦ : f(x) 6= y.

Then

{f(I ′′◦) ⊆ (−∞, y)⇒ f(I ′′−) ⊆ (−∞, y]f(I ′′◦) ⊆ (y,+∞)⇒ f(I ′′−) ⊆ [y,+∞)

, since f is a Darboux function.

If ∃I◦0 ⊆ I◦ : f(I◦0 ) ⊆ (0,+∞), we consider a sequence (I◦n : n ∈ N) such thatI◦n ⊆ I◦n−1 : f(I◦n) ⊆ (n,+∞) for all n ∈ N∗. Then (Theorem 3) there is a certain I−∞

such that I−∞ =⋂n∈N

I−n ⇒ f(I−∞) = f

(⋂n∈N

I−n

)⊆⋂n∈N

f(I−n ) ⊆⋂n∈N

[n,+∞) = ∅,

a contradiction, so there is a certain n0 ∈ N∗ such that f(I◦n0) ⊆ (−∞, n0). Then

f(I◦n0) = f(I◦0 ∩ I◦n0

) ⊆ f(I◦0 ) ∩ f(I◦n0) ⊆ (0,+∞) ∩ (−∞, n0) = (0, n0).

We now consider three sequences (yn) ∈ RN, (I◦yn: n ∈ N), (J◦n : n ∈ N) such

that y0 = n0, J◦0 = (0, n0) and yn = 12 (sup J◦n−1 + inf J◦n−1),

I◦yn⊆ I◦yn−1

:

{f(I◦yn

) ⊆ (−∞, yn)f(I◦yn

) ⊆ (yn,+∞), J◦n =

{(inf J◦n−1, yn), if f(I◦yn

) ⊆ (−∞, yn)(yn, sup J◦n−1), if f(I◦yn

) ⊆ (yn,+∞)

for all n ∈ N∗. Then f(I◦yn) = f(I◦yn−1

∩ I◦yn) ⊆ f(I◦yn−1

) ∩ f(I◦yn) ⊆ J◦n for all

n ∈ N∗ and f(I◦y0) ⊆ J◦0 by definition, so f(I−yn

) ⊆ J−n for all n ∈ N, since f is a

Darboux function, as well as limn→+∞

(max J−n − min J−n ) = limn→+∞

12nn0 = 0. Then

(Theorem 3) there are certain I−y∞ and ξ ∈ R such that

I−y∞ =⋂n∈N

I−yn⇒ f(I−y∞) = f

(⋂n∈N

I−yn

)⊆⋂n∈N

f(I−yn) ⊆

⋂n∈N

J−n = {ξ}.

It cannot be f(I−y∞) = ∅, so f(I−y∞) = {ξ}. This yields that f is continuous at I◦y∞ ,but this is not yet a contradiction, since I◦y∞ can be empty. It can be shown thatthe progressively tighter bounding of the image of open intervals that are supersetsof I−y∞ around ξ results in f being continuous at ∂I−y∞ as well. We conclude that f

is continuous at I−y∞ , a contradiction. Likewise if ∃I◦0 ⊆ I◦ : f(I◦0 ) ⊆ (−∞, 0). So

there are certain ζ ∈ R and I ′◦ ⊆ I◦ such that f−1({ζ}) is dense in I ′◦.Let f−1({y}) be nowhere-dense in I ′◦ for all y ∈ R \ {ζ}. This means that

∀y ∈ R : ∀I ′′◦ ⊆ I ′◦ : ∃I ′′′◦ ⊆ I ′′◦ : ∀x ∈ I ′′′◦ : f(x) 6= y.

Then f(I ′′′◦) ⊆{

(y,+∞), if y ∈ (−∞, ζ)(−∞, y), if y ∈ (ζ,+∞)

, since f is a Darboux function and

f−1({ζ}) is dense in I ′◦.We consider two sequences (yn) ∈ RN, (I◦yn

: n ∈ N) such that y0 = ζ + 1,I◦y0⊆ I ′◦ : f(I◦y0

) ⊆ (−∞, y0)

and

{yn = ζ + 1/2n, I◦n ⊆ I◦n−1 : f(I◦yn

) ⊆ (−∞, yn), if n is evenyn = ζ − 1/2n, I◦n ⊆ I◦n−1 : f(I◦yn

) ⊆ (yn,+∞), if n is oddfor all n ∈ N∗.

Then (Theorem 3) there is a certain I−y∞ such that I−y∞ =⋂n∈N

I−yn⇒

⇒ f(I−y∞) = f

(⋂n∈N

I−yn

)⊆⋂n∈N

f(I−yn) ⊆

⋂n∈N

{(−∞, yn] , if n is even[yn,+∞) , if n is odd

= {ζ}

4 SPYRIDON TZIMAS

It cannot be f(I−y∞) = ∅, so f(I−y∞) = {ζ}. Using the same argument as above,

we conclude that f is continuous at I−y∞ , a contradiction. So there are certain

η ∈ R \ {ζ} and I ′′◦ ⊆ I ′◦ such that f−1({η}) is dense in I ′′◦. Therefore, f−1({y})is dense in I ′′◦ for all y ∈ [min{ζ, η},max{ζ, η}], since f is a Darboux function. �

Theorem 5. Let f : I◦ → R be discontinuous everywhere. Then f does nothave an antiderivative.

Proof. Let f have an antiderivative. Then (Theorem 1) f is also a Darbouxfunction. Then (Theorem 4) f |I ′′◦ is a Conway function for a certain I ′′◦ ⊆ I◦, so(Theorem 2) f |I ′′◦ does not have an antiderivative, a contradiction. �

Theorem 6. Let f : I◦ → R have an antiderivative. Then f is continuous ata dense subset of I◦.

Proof. Let f not be continuous at a dense subset of I◦. Then ∃I ′◦ ⊆ I◦ : fis discontinuous at I ′◦, so (Theorem 5) f |I ′◦ does not have an antiderivative. �

Definition. A set S ⊆ R is a meager set, if and only if it is the union of acountable collection of nowhere-dense subsets of R.

Theorem 7. Let f : I◦ → R be a function that is continuous at a dense subsetof I◦. Then f is discontinuous at a meager subset of I◦.

Proof. We defineXc to be the set of points where f is continuous andXd to bethe set of points where it is discontinuous. Moreover, xc ∈ Xc and xd ∈ Xd. Then∀xc : ∀ε > 0 : ∃δ(xc, ε) > 0 : ∀x ∈ I◦∩(xc−δ(ε, xc), xc+δ(ε, xc)) : |f(x)−f(xc)| < ε.

We define A(ε) =⋃xc

I◦ ∩ (xc − δ(ε, xc), xc + δ(ε, xc)) for all ε > 0.

Then A(ε) ⊇ Xc ⇒ I◦ ∩ AC(ε) ⊆ Xd. If I◦ ∩ AC(ε) is the empty set or a unitset, then it is nowhere-dense. Let it be neither and be dense at a certain I ′◦ ⊆ I◦.Then ∃xc ∈ I ′◦ ∩ A(ε) : ∃xd ∈ I ′◦ ∩ AC(ε) : xd ∈ I ′◦ ∩ (xc − δ(ε, xc), xc + δ(ε, xc)),a contradiction, so I◦ ∩AC(ε) is nowhere-dense. Furthermore,⋂

n∈N∗A( 1

n ) = Xc ⇒⋃

n∈N∗I◦ ∩AC( 1

n ) = Xd,

so Xd is the union of a countable collection of nowhere-dense sets - a meager set. �

Theorem 8. Let f : I◦ → R be a function that has an antiderivative. Then fis discontinuous at a meager subset of I◦.

Proof. Let f have an antiderivative. Then (Theorem 6) f is continuous at adense subset of I◦, so (Theorem 7) it is discontinuous at a meager subset of I◦. �

References

[1] http://en.wikipedia.org/wiki/Smooth function#Examples

[2] http://en.wikipedia.org/wiki/Nowhere continuous function

[3] http://en.wikipedia.org/wiki/Conway base 13 function

The Author. Spyridon Tzimas is an undergraduate student at the Depart-ment of Mathematics of the University of Ioannina, Greece. His list of researchinterests still includes any and all open problems in Mathematics, but it will changeto something less exhaustive once he gets his degree. Or maybe it won’t.

E-mail address: ma09252@cc.uoi.gr