I have a question from Munkres' Analysis on Manifolds textbook, part of which i am having trouble with.
Theorem: Let $S$ be a bounded set in $\mathbb R^n$; let $f\colon S\to\mathbb R$ be a bounded function. Let $D$ be the set of points of S at which f fails to be continuous. Let E be the set of points of Bd S (boundary of S) at which the condition lim f(x)=0 as x approaches $x_0$, fails to hold. Then the integral of f over S exists if and only if D and E have measure zero.
Proof. (a) Show that f restricted to S is continuous at each point x_0 not in being in the set (D union E)
b) Let B be the set of isolated points of S; then B subset of E because the limit cannot be defined if x_0 is not a limit point of S. Show that if F restrict to S is continuous at x_0, then x_0 is not an element of the set D union (E\B).
c) Show that B is countable.
d) complete the proof.
I am having trouble doing part c). For some reasons, i don't know how to show the set of isolated points are countable.
For any set $S\subset \mathbb R^n$ the set of isolated points of $S$ is [at most] countable. Indeed, if $x$ is an isolated point, then [by the definition of an isolated point] there is $r_x>0$ such that the ball $B(x,r_x)$ does not intersect $S\setminus \{x\}$. The open balls $B(x,\frac12 r_x)$ are pairwise disjoint (check this). Since each of them contains a point with rational coordinates, there can be at most countably many such balls; hence at most countably many isolated points $x$.