Types of Discontinuities

4.4 Types of Discontinuities

When you have eliminated the impossible, whatever remains, however improbable, must be the truth.

Sir Arthur Conan Doyle (1859–1930), The Sign of Four

We start with a basic property that asserts that monotone functions do not have discontinuity points of the second kind.

4.4.1. Let I be a nondegenerate interval and assume that f :I →R is monotone. Prove that if f has discontinuity points, then they are of the first kind, and the set of discontinuity points is at most countable.

Solution. Assume that f is nondecreasing. Let a ∈ I be such that lim x → a − f (x) makes sense. We claim that

a (x) = sup f → (x) =: ℓ . − x <a

lim f

Indeed, since f (x) ≤ f (a) for all x < a, it follows that ℓ < +∞. Fix ε > 0. Thus, there exists x ε ∈ I, x ε < a, such that ℓ − ε < f (x ε ) ≤ ℓ. Since f is nondecreasing, we obtain

ℓ− ε < f (x) ≤ ℓ for all x ∈ [x ε , a),

152 4 Continuity which shows that ℓ = lim x → a − f (x). We treat analogously the case of points a ∈ I

such that lim x → a + f (x) makes sense. If a is an interior point of I, then both one- sided limits make sense, and moreover, lim x → a − f (x) ≤ lim x → a + f (x).

Let D denote the set of discontinuity points of f . We associate to any a ∈ D the nondegenerate interval

lim f (x), lim f (x) .

If one of these limits does not make sense, then we replace it by f (a) in the definition

of I a . We observe that if a, b ∈ D, then I a ∩I b

to any a ∈ D we can associate r a ∈I a ∩ Q; hence we define a one-to-one mapping

D a ∈ Q. Since Q is countable, it follows that D is at most countable.

A deeper property established in 1929 by the Romanian mathematician Alexandru Froda [28] asserts that the set of discontinuity points of the first kind of any function f : R →R is at most countable. We refer to [82] for a detailed proof of this result.

Consider the Riemann function f : R →R defined by

0 if x = 0 or if x ∈ R \ Q,

f (x) = 1

if x = m n , m ∈ Z,n ∈ N ∗ , (m, n) = 1. Then f is continuous on (R \ Q) ∪ {0} and discontinuous on Q \ {0}. A natural

question is whether there is a function f : R →R that is continuous at rational points and discontinuous at irrational points. The answer to this question is no. This is

a direct consequence of the following general result, which is due to the Italian mathematician Vito Volterra (1860–1940), based on the observation that both Q and R \ Q are dense subsets of R.

In 1898, Osgood proved the following interesting property of the real axis: if (U n ) n ≥1 is a sequence of open and dense subsets in R, then their intersection is dense in R, too. In the more general framework of complete metric spaces this result is known as Baire’s lemma, after Baire, who gave a proof in 1899.

A nice application of this property was found in 1931 by Banach, who showed that “almost” continuous functions f : [0, 1]→R do not have a right derivative at any point. Indeed, let us observe that the space of all continuous functions on [0, 1] (denoted by C [0, 1]) becomes a complete metric space with respect to the distance

d ( f , g) = max x ∈[0,1] | f (x) − g(x)|, for all f ,g ∈ C[0,1]. For any positive integer n, let F n denote the set of all functions f : [0, 1]→R for which there exists some x (depending on f ) such that for all x ≤x ′ ≤ x + 1/(n + 1),

0 ≤x≤1− and | f (x ′ ) − f (x)| ≤ n(x ′

− x).

Then the set F n is closed in C [0, 1], so its complementary set U n is open. Moreover, U n is dense in C [0, 1]. This follows from the fact that for any f ∈ C[0,1] and for all M > 0 and ε > 0, there exists g ∈ C[0,1] such that d( f ,g) < ε , and the right

4.4 Types of Discontinuities 153 derivative of g exists in any point, and moreover, its absolute value is at least M.

For this purpose it is enough to choose a curve such as the “sawtooth,” whose graph lies in a tape of width ε around the graph of f , and the number of “teeth” is suffi- ciently large that any segment of the graph has a slope that is either greater than M or less than −M. At this stage, by Baire’s lemma, it follows that the set U

= n ≥1 U n is dense in C [0, 1]. Using now the definition of F n , it follows that no function f ∈ U can admit a finite right derivative at any point in [0,1]. ⊓ ⊔

We do not insist in this direction, but we will prove a property in the same dir- ection as that of Osgood. More precisely, Volterra [113] proved in 1881 that if two real continuous functions defined on the real axis are continuous on dense subsets of R, then the set of their common continuity points is dense in R, too. The fol- lowing result is a generalization of Volterra’s theorem in the sense that it asserts, additionally, that the set of common continuity points is uncountable.

4.4.2. Consider the functions f , g : R→R and denote the set of continuity points of f (resp. g ) by C f (resp. C g ). Assume that both C f and C g are dense in R . Prove that C f ∩C g is an uncountable dense subset of R .

Solution. We first observe that it is enough to prove that for any real a and for all r > 0,

f ∩C g (4.1) Indeed, by (4.1), (a − r,a + r) ∩ C f ∩C g f ∩C g is a dense subset of R.

Relation (4.1) also implies that the set C f ∩C g is uncountable. Indeed, if not, we choose C =C f ∩C g , so

f ∩C g

(4.1). For any fixed a ∈ R and r > 0, set a 0 = a and r 0 = r. Define inductively the sequences (a n ) n ≥0 ⊂ E and (r n ) n ≥0 ⊂R + as follows:

(i) [a n +1 −r n +1 ,a n +1 −r n +1 ] ⊂ (a n −r n ,a n +r n ), for all positive integers n; (ii) c n

n −r n ,a n +r n ), for all n ≥ 1; (iii) if n is odd, then | f (x) − f (y)| < 1/n, for all x,y ∈ (a n −r n ,a n +r n ); (iv) if n ≥ 2 is even, then |g(x) − g(y)| < 1/n, for all x,y ∈ (a n −r n ,a n +r n ).

These sequences are constructed as follows: assume that we have defined a k and r k , for all k < n. If n is odd, we choose a n ∈ (a n −1 −r n −1 ,a n −1 +r n −1 )∩C f . We observe that such an element exists, by the density of C f in R. Using now the con- tinuity of f in a n , there exists δ > 0 such that | f (x) − f (a n )| < 1/2n, provided |x − a n |< δ . Choosing now r n > 0 such that r n < min{ δ , r/n, r n −1 − |a n −a n −1 |}, we deduce that if x , y ∈ (a n −r n ,a n +r n ), then

| f (x) − f (y)| ≤ | f (x) − f (a n )| + | f (a n ) − f (y)| < . n

We also observe that [a n −r n ,a n +r n ] ⊂ (a n −1 −r n −1 ,a n −1 +r n −1 ). Indeed, if |x − a n |≤r n , then

|x − a n −1 | ≤ |x − a n | + |a n −a n −1 |≤r n + |a n −a n −1 |<r n −1 .

154 4 Continuity If n is even, we define similarly a n and r n , with the only observation that we

replace f (resp. C f ) by g (resp. C g ).

By (i) and since (r n ) converges to zero, it follows by Cantor’s principle that there exists b

∈ n ≥0 [a n −r n ,a n +r n ]. Moreover, (i) also implies that b ∈ B(a n ,r n ), for all n ≥ 0.

We prove in what follows that f and g are continuous in b, that is,

b ∈C f ∩C g . Fix ε > 0 and an odd integer N such that 1/N < ε . By (iii), it follows that if δ <r N − |b − a N | then | f (x) − f (b)| < 1/N < ε , provided that |x − b| < δ , so

f is continuous in b. A similar argument shows that b ∈C g . Since b ∈ (a n −r n ,a n +r n ) and for all positive integers n, c n

n −r n ,a n +r n ), it follows that b