Here’s a pair of facts I hadn’t much considered before the last time I taught real analysis:
A monotone function can have at most countably many discontinuities.
One thinks immediately of the floor function f(x)=⌊x⌋, which has a jump of size 1 at every integer. It is possible to have infinitely many jumps within a finite interval; for instance, 1⌊1/x⌋ has a jump at every unit fraction 1/n. But can a monotone function have a dense set of discontinuities, say, the set of rationals? Sure, here’s one:
That’s the graph of f(x)=∞∑n=1⌊nx⌋2n. It has a jump of size 1/(2q−1) at each fraction of the form p/q (when p/q is in reduced form, of course). Exercise. Why are these the sizes of the jumps? Keep in mind where the discontinuities of ⌊nx⌋ appear.Here’s a general way to construct a monotone function f:R→R whose set of discontinuities is your favorite countable set C. Let c1, c2, c3, … be an enumeration of C, and for each x∈R, set N(x)={n∈N:cn≤x} Let a1, a2, a3, … be a sequence of strictly positive numbers such that ∑an<∞. Then define f(x)=∑n∈N(x)an. This function is discontinuous at each point in C and continuous at each point that is not in C. (Exercise. Why? Keep in mind that the tail of a convergent series can be made arbitrarily small.) Essentially, we have created a distribution by placing a “delta mass” of weight an at the point cn, and f is the integral of this distribution from −∞ to x.
When C is dense, the construction above produces a strictly increasing function f, and moreover the image of f is nowhere dense. I call the graph of such a function an “angels’ staircase”, because f is a one-sided inverse of a “devil’s staircase” function g—that is, g is a monotone increasing function that is constant except on a set of measure zero. (MathWorld, on the other hand, uses “devil’s staircase” to refer to both kinds of functions.)
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