An increasing function on a closed interval has countable many points of discontinuity. [duplicate]

An exercise from "Mathematical Analysis" of T. M. Apostol.

Let $$f$$ be an increasing function defined on $$[a, b]$$ and let $$x_1, ... , x_n$$ be $$n$$ points in the interior such that $$a < x_1 < x_2 < ... < x_n < b$$.

a) Show that $$\sum_{k=1}^n [f(x_k+) - f(x_k-)] \leq f(b-) - f(a+)$$.

b) Deduce from part (a) that the set of discontinuities of $$f$$ is countable.

c) Prove that $$f$$ has points of continuity in every open subinterval of $$[a, b]$$.

Here $$f(x-)$$ and $$f(x+)$$ mean the left-hand and right-hand limit of $$f$$ in $$x$$.

Solution for (a)

For (a), by $$f$$ increasing, we have $$f(b') - f(x_n) + \sum_{k=1}^{n-1} [f(x_{k+1}) - f(x_k)] = f(b') - f(x_1) \leq f(b') - f(a')$$ for every $$a' \in (a,x_1)$$ and $$b' \in (x_n, b)$$. Hence taking the supremum of $$f(b')$$ for $$b' \in (x_n, b)$$ and the infimum of $$f(a')$$ for $$a' \in (a,x_1)$$, calling $$b'=x_{n+1}$$, we have:

$$\sum_{k=1}^{n} [f(x_{k+1}) - f(x_k)] \leq f(b-) - f(a+)$$

Notice that $$f(b-)$$ must exist and can't be infinity because $$f(b') \leq f(b-) \leq f(b)$$, the same for $$f(a+)$$.

For $$f$$ is increasing, $$f(x_{k+1}) \geq f(x_k+)$$ and $$f(x_{k}) \geq f(x_k-)$$ hence $$f(x_k+) - f(x_k-) \leq f(x_{k+1}) - f(x_{k})$$, and we have:

$$\sum_{k=1}^{n} [f(x_k+) - f(x_k-)] \leq f(b-) - f(a+)$$

Solution for (c)

From (b) is trivial, if $$f$$ is disconitnuous everywhere in an open subinterval of $$[a, b]$$ the $$f$$ has uncountable many points of discontinuity.

I'm asking for a solution for (b).

• I saw that question but i didn't figure immediately how to translate the result in this settings. Apr 24, 2020 at 9:49
• Another one: math.stackexchange.com/q/84870/42969. Apr 24, 2020 at 9:51

Fix a positive integer $$N$$. Suppose there are $$n$$ points $$x_k, 1 \leq k\leq n$$ with $$f(x_k+)-f(x_k-) \geq \frac 1 N$$. Then a)gives $$\frac n N \leq C$$ where $$C=f(b-)-f(a+)$$. Hence there are at most $$NC$$ such points. Note that the set of points where $$f$$ is not continuous is the union of these points over all $$N$$ (because $$f(x+)-f(x-) >0$$ iff $$f(x+)-f(x-) \geq \frac 1 N$$ for some $$N$$). Since countable union of finite sets is countable we are done.