This is the problem from Munkres's Analysis of manifold 13.7. I'm not sure about this problem so I want to verify my ideas about the first 3 parts of this question. Then I hope there can be some help with the last part.
For the notation: $$f_S(x)= \begin{cases} f(x) & x\in S \\ 0 & x\notin S\end{cases}$$
For (a) I think that $ D=\{x_0\in S: f \text{ is not continuous at } x_0\} $ and $E=\{x_0\in Bd(S):\lim _{x\to x_0 }f(x)\ne 0\}$
And $x_0\notin D\cup E\implies$ $$x_0\in D^c\cap E^c=\{x_0\in S\cap Bd(S): \ f \text{ is continuous at }x_0 \ and \lim _{x\to x_0 }f(x)= 0\}$$
Then I need to verify that for each $x_0$ $f_S$ is continuous.
For (b) I think I can show the equivalent negation which is $$x_0\in D\cup(E-B)\implies f_S \text{ is not continuous at } x_0$$
and it's sufficient to only prove
$1.\ x\in D$ and $x\notin E-B$ ; and $2. \ x\notin D$ and $x\in E-B$
For $1.$, since $f $ is not continuous on $x_0$, it's clear that $f_S$ won't be continuous on it
For $2.$, since $x_0\in Bd(S)$, for $x\in int(S^c)$ $\lim_{x\to x_0}f_S(x)=0$. However, since $x_0 $ is a limit point, and for $x\in int(S)$, $\lim_{x\to x_0} f_S(x)=\lim_{x\to x_0}f(x)\ne 0$. Thus the limit of $f_S$ even doesn't exist, so $f_S$ is not continuous for each point in $E-B$
For (c)
Since each point $x\in B$ is an isolated point, for each $x$, I can choose an open rectangle $ I_x=(a_1^x,b_1^x)\times \cdots \times (a_n^x, b_n^x)$ where each $a_1^x,b_1^x,\cdots, a_n^x, b_n^x$ is a rational number such that $I_x\cap S=\{x\}$. And $(a_1^x,b_1^x,\cdots, a_n^x, b_n^x)$ can be regarded as a point in $\mathbb Q^{2n}$. Also, each $x\in B$ is corresponding to a unique rectangle(point in $\mathbb Q^{2n}$) After some work, I can say there exists an injection $i: B\to \mathbb N$, so $B$ is countable
For (d)
Suppose $\displaystyle \int_S f$ exists, then let $I$ be any rectangle such that $int(I)\supset \overline{S}$, and $\displaystyle \int_I f_S$ exists. Then, since $\mathbb R^n =\overline S \cup (\overline S)^c$, and $f_S\equiv 0 $ on $(\overline S)^c$. Thus $f_S$ is only discontinuous on $\overline S= S\cup Bd(S)$
Discontinuous point set for $f_S$ on $S$ is just discontinuous point set for $f$ which is $D$ since $f=f_S$ on $S$, so since $\displaystyle \int_S f$ exists, $D$ it has measure zero. And for points $x_0$ in $Bd(S)$, $f_S$ is discontinuous at $x_0$ where $\lim_{x\to x_0}f(x)\ne 0$ which is exactly $E$. Thus $E$ has measure zero.
Then for the other side. If $D, E$ has measure zero, then since they are the only set of discontinuous points of $f_S$, $\displaystyle \int_I f_S$ exists and so $\displaystyle \int_S f$ exists.
I think there must be some problems here since, in the last part, I didn't use anything brought by parts (b) and (c). Maybe there are flaws in parts a,b, and c of my attempt. Any help on this? Thanks!
