Discontinuities of a real function

Let f ⁣:RRf\colon\R\to\R be a real function. Then, ff is continuous at aa if and only if for all ε>0,\varepsilon>0, there exists a δ>0\delta>0 so that f(x)f(a)<ε|f(x)-f(a)|<\varepsilon for all xx with xa<δ.|x-a|<\delta. Our task today is to investigate the discontinuities of f,f, i.e. the set D(f)D(f) of points at which ff is not continuous. Specifically, we want to answer two questions:

  1. What type of set can D(f)D(f) be?

  2. Given any set A,A, is there a function ff that is discontinuous exactly on AA?

Back and forth

We first need to know what it means for ff to be discontinuous at a point. By negating the definition of continuity, we can deduce that

ff is discontinuous at aa if and only if there exists an ε>0\varepsilon>0 such that for all δ>0,\delta>0, we have f(x)f(a)ε|f(x)-f(a)|\ge\varepsilon for some xx with xa<δ.|x-a|<\delta. (*)

The definition should make sense to you: it says that as we zoom in closer to a,a, the function ff stops getting closer to f(a)f(a) at some point, instead staying outside a “barrier” around f(a).f(a).

Given a set I,I, the oscillation of ff on II is

ω(f;I)=sup{f(x)f(y) ⁣:x,yI}.\omega(f;I)=\sup\{ |f(x)-f(y)|\colon x,y\in I \}.

The geometric meaning of the oscillation is hopefully clear. Note that oscillations are always nonnegative, and if JJ is a subset of I,I, then ω(f;J)ω(f;I).\omega(f;J)\le\omega(f;I). Thus, as II “shrinks” toward a,a, the oscillation ω(f;I)\omega(f;I) decreases to some nonnegative limit. We can now define the oscillation of ff at aa by

ωf(a)=limh0+ω(f;(ah,a+h)).\omega_f(a) = \lim_{h\to 0^+} \omega(f; (a-h,a+h)).

If ω(f;I)\omega(f;I) is how much ff “jumps” on the set I,I, then ωf(a)\omega_f(a) is how much ff jumps at a.a. The definition (*) above can be rewritten into

ff is discontinuous at aa if and only if there exists an ε>0\varepsilon>0 such that for all δ>0,\delta>0, we have ωf(x,(aδ,a+δ))ε.\omega_f(x, (a-\delta,a+\delta))\ge\varepsilon.

We can now conclude that ff is discontinuous at aa if and only if ωf(a)>0.\omega_f(a)>0.

Open sesame

The preceding observation allows us to write D(f)D(f) as the set of aa for which ωf(a)>0.\omega_f(a)>0. We need one final detail.

Lemma. For any r>0,r>0, the set {a ⁣:ωf(a)<r}\{a\colon\omega_f(a)< r\} is open.

Proof. Given aa such that ωf(a)<r,\omega_f(a)<r, there is a bounded interval II containing aa such that ω(f;I)<r.\omega(f;I)<r. Consequently, II is a subset of {a ⁣:ωf(a)<r},\{a\colon\omega_f(a)< r\}, since for any xI,x\in I, we have ωf(x)ω(f;I)<r.\omega_f(x)\le\omega(f; I)<r. \square

With this lemma, we can finally prove a characterization of D(f).D(f).

Theorem. D(f)D(f) is a countable union of closed sets in R.\R.

Proof. The first part of this characterization is easy enough:

D(f)={a ⁣:ωf(a)>0}=n=1{a ⁣:ωf(a)1/n}.\begin{align*} D(f) &= \{a\colon \omega_f(a) > 0\} \\ &= \bigcup_{n=1}^\infty \{a\colon \omega_f(a)\ge 1/n\}. \end{align*}

The complement of {a ⁣:ωf(a)1/n}\{a\colon \omega_f(a)\ge 1/n\} is {a ⁣:ωf(a)<1/n},\{a\colon \omega_f(a)< 1/n\}, which is open. Thus, D(f)D(f) is a countable union of closed sets. \square

This result holds in a general metric space. In a topological space, a countable union of closed sets is called an FσF_\sigma set.1 We now know that D(f)D(f) is an FσF_\sigma set, but is every FσF_\sigma set a D(f)D(f) for some ff?

The imitation game

The answer to the question above is “yes”: any FσF_\sigma set in R\R is the set of discontinuities for some function f.f. We shall construct ff by trying to follow the characterization in our proof of the major theorem in the previous section.

Theorem. If EE is an FσF_\sigma set in R,\R, there is some f ⁣:RRf\colon\R\to\R such that E=D(f).E=D(f).

Proof. Let E=i=1Fi,E=\bigcup_{i=1}^\infty F_i, where each FiF_i is closed. Let E1=F1E_1=F_1 and

En=(i=1nFi)(i=1n1Fi).E_n = \left(\bigcup_{i=1}^n F_i\right) \setminus \left(\bigcup_{i=1}^{n-1} F_i\right).

Note that the EnE_n are disjoint, n=1mEn=n=1mFn,\bigcup_{n=1}^m E_n = \bigcup_{n=1}^m F_n, and n=1En=E.\bigcup_{n=1}^\infty E_n = E. Each EnE_n has a countable dense subset DnD_n (see later). We now define ff on R\R as follows:

f(x)={0xRE,2n+1xDn,2nxEnDn.f(x) = \begin{cases} 0 & x \in \R\setminus E, \\ 2^{-n+1} & x \in D_n, \\ 2^{-n} & x\in E_n\setminus D_n. \end{cases}

The function is well-defined since RE\R\setminus E and the EnE_n are mutually disjoint. We now consider some xR.x\in\R.

Suppose xE,x\in E, then xEnx\in E_n for some n.n. Consider two possibilities:

In any case, we can always find a point yy in this interval with f(x)f(y)2n.|f(x)-f(y)|\ge 2^{-n}. Altogether, if xEn,x\in E_n, then ωf(x)2n>0,\omega_f(x)\ge 2^{-n}>0, so ff is discontinuous at x.x.

Suppose xRE.x\in \R\setminus E. For all integer m>0,m>0, the set Am=n=1mEn=n=1mFnA_m=\bigcup_{n=1}^m E_n=\bigcup_{n=1}^m F_n is closed and does not contain x.x. For any positive integer N,N, RAN\R\setminus A_N is open, so there is an open interval (xr,x+r)(x-r,x+r) disjoint from AN,A_N, in which case the interval only contains points in RE\R\setminus E or in EkE_k with k>N.k>N. Hence, for any yy with xy<r,|x-y|<r, we have f(x)f(y)2N,|f(x)-f(y)|\le 2^{-N}, and since NN is arbitrary, this means the oscillation of ff at xx is zero. Thus, ff is continuous at x,x, and we conclude D(f)=E.D(f)=E. \square

There is one detail we need to work out, and that is each EnE_n has a countable dense subset, i.e. it is separable. Luckily for us, a subset of a separable metric space is always separable, and R\R is indeed separable, so EnE_n is separable too. This detail will be left an exercise for the reader. After that, it should not be difficult to see that this theorem holds for any separable metric space, not just the reals, and the proof is mostly analogous.

To recap, we answer the two questions put forth at the beginning of the post:

  1. What type of set can D(f)D(f) be?

    It must be a countable union of closed sets; in other words, an FσF_\sigma set.

  2. Given any set A,A, is there a function ff that is discontinuous exactly on AA?

    In a separable metric space M,M, for any FσF_\sigma set, there is a function f ⁣:MMf\colon M\to M whose D(f)D(f) is that set.


  1. FF here stands for the French word fermé, meaning closed, while the σ\sigma means a “sum” of sets, i.e. a union.