What shown below is a reference from "Analysis on manifolds" by James R. Munkres
Unfortunately I can't prove the last theorem as suggested by Munkres. Anyway it seems to me that I found another way to prove it so I ask if it is correct and if not I ask to prove the last theorem.
Let be $Q$ a rectangle that contains $S$ and for convenience we call $K$ the set of the points where $f_S$ is continuous and we prove that $K=(D\cup E)^C$. First of all we observe that $D\subseteq S$ and $E\subseteq\partial S$ thus $D\cup E\subseteq S\cup\partial S=\overline{S}$.
So for $x_0\notin D\cup E$ we analyse the case where $x_0\in\overline{S}$ and the case where $x_0\notin\overline{S}$ and so we prove that the function $f_S$ is continuous in $x_0$.
- So if $x_0\in\overline{S}$ then or $x_0\in\overset{°}S$ or $x_0\in\partial S$. So if $x_0\in\overset{°}S$ then $f_S(x_0)=f(x_0)$ and so $f_S$ is continuous in $x_0$, since $x_0\notin D$ and $f_S(x)=f(x)$ for any $x\in\overset{°}S$ so that by continuity of $f$ in $x_0$ and by the uniqueness of the limit in hausdorf spaces we argue that for any succession $x_n$ converging to $x_0$ it is $$\underset{x\rightarrow x_0}{\text{lim}}f_S(x)=\underset{n\rightarrow+\infty}{\text{lim}}f_S(x_n)=\underset{n\rightarrow+\infty}{\text{lim}}f(x_n)=\underset{x\rightarrow x_0}{\text{lim}}f(x)=f(x_0)=f_S(x_0)$$ because if $x_n\rightarrow x_0$ then there exist $n_0\in\Bbb{N}$ such that $x_n\in\overset{°}S$ for any $n\ge n_0$ and so $\underset{n\rightarrow+\infty}{\text{lim}}f_S(x_n)=\underset{n\rightarrow+\infty}{\text{lim}}f(x_n)$. Instead if $x_0\in\partial S$ then by hypothesis $\underset{x\rightarrow x_0}{\text{lim}}=0$ and so for any $\epsilon>0$ there exist $\delta>0$ such that $|f(x)-0|<\epsilon$ for any $x\in C(x_0,\delta)\cap S$ so that if we prove that $f_S(x_0)=0$ then $$|f_S(x)-f_S(x_0)|<\epsilon$$ for any $x\in C(x_0,\delta)$, since if $x\in C(x_0,\delta)\cap S$ then $|f_S(x)-f_S(x_0)|=|f(x)-0|<\epsilon$ and if $x\in C(x_0,\delta)\cap S^C$ then $|f_S(x)-f_S(x_0)|=0<\epsilon$. So we observe that if $x_0\in S$ then the function $f$ is here continuous and so $0=\underset{x\rightarrow x_0}{\text{lim}}f(x)=f(x_0)=:f_S(x_0)$ and if $x_0\notin S$ then $f_S(x_0):=0$.
- If $x_0\notin\overline{S}$ then $x_0\in\text{ext}(S)$ and so for some neighborhood $V_{x_0}$ of the neighborhood system $\mathcal{V}(x_0)$ the function $f_S$ agree with zero function $0_{\Bbb{R}^n}$ that obviously is here continuous so that by this and by the uniqueness of the limit in hausdorff space for any succession $x_n$ of points of $S$ converging to $x_0$ we argue that $$0=\underset{n\rightarrow+\infty}{\text{lim}}0_{\Bbb{R}^n}(x_n)=\underset{n\rightarrow+\infty}{\text{lim}}f_S(x_n)=\underset{x\rightarrow x_0}{\text{lim}}f_S(x)$$ and so the function $f_S$ is continuous in $x_0$.
So we have proved that if $x_0\notin D\cup E$ then $x_0\in K$ and so $(D\cup E)^C\subseteq K$.
Now we prove that if $x_0\in D\cup E$ then $f_S$ is discontinuous. So for what we initially observed if $x_0\in D\cup E$ then $x_0\in\overline{S}$ and so or $x\in\overset{°}S$ or $x\in\partial S$ thus we distinguish this two cases.
- So if $x_0\in\overset{°}S$ then $f_S(x_0)=f(x_0)$ and so $f_S$ is discontinuous at $x_0$, since $x_0\in D$ and $f_S(x)=f(x)$ for any $x\in\overset{°}S$ so that by the discontinuity of $f$ in $x_0$ we argue that there exist a succession $x_n$ converging to $x_0$ such that $f(x_0)\neq\underset{x\rightarrow x_0}{\text{lim}}f(x_n)$ and so we arghe that $$f_S(x_0)\neq\underset{x\rightarrow x_0}{\text{lim}}f_S(x)$$ since $f_S(x_0)=f(x_0)\neq\underset{n\rightarrow+\infty}{\text{lim}}f(x_n)=\underset{n\rightarrow+\infty}{\text{lim}}f_S(x_n)$.
- Instead if $x_0\in\partial S$ we observe that any succession $x_n$ of points of $S$ converging to $x_0$ is such that $\underset{n\rightarrow+\infty}{\text{lim}}f(x_n)=\underset{n\rightarrow+\infty}{\text{lim}}f_S(x_n)$ so that if $\underset{x\rightarrow x_0}{\text{lim}}f_S(x)$ exist then $\underset{x\rightarrow x_0}{\text{lim}}f(x)$ exist too and so we observe that if $x_0\in\partial S$ then $x_0\in\overline{S}$ and $x_0\in\overline{S^C}$ so that by the first countability of $\Bbb{R}^n$ there exist a succession $y_n$ in $S$ and a succession $z_n$ in $S^C$ converging to $x_0$ thus, since $\underset{n\rightarrow+\infty}{\text{lim}}f_S(y_n)=\underset{n\rightarrow+\infty}{\text{lim}}f(y_n)$ and $\underset{n\rightarrow +\infty}{\text{lim}}f_S(z_n)=\underset{n\rightarrow+\infty}{\text{lim}}0_{\Bbb{R}^n}(z_n)=0$, if the function $f_S$ was continuous at $x_0$ then $f_S(x_0)=\underset{x\rightarrow x_0}{\text{lim}}f_S(x)$ and for the uniqueness of the limit in hausdorff spaces it would be $\underset{x\rightarrow x_0}{\text{lim}}f_S(x)=\underset{n\rightarrow+\infty}{\text{lim}}f_S(z_n)=0$ thus if $x_0\in\partial S$ then $f_S$ is discontinuous at $x_0$ because $x_0\in E$ and so $\underset{x\rightarrow x_0}{\text{lim}}f_S(x)=\underset{n\rightarrow+\infty}{\text{lim}}f_S(y_n)\neq 0$.
So we conclude that if the function $f_S$ is continuous in $x_0$ then $x_0\notin D\cup E$ and so $K\subseteq(D\cup E)^C$.
So we have proved that $K=(D\cup E)^C$ and so the set of the points where $f_S$ is discontinuous is $D\cup E$.
So if $\int_Sf$ exist then $D\cup E$ has measure zero and so even $E$ and $D$ have measure zero, because $E,D\subseteq D\cup E$. Viceversa if $D$ and $E$ have measure zero then $E\cup D$ has measure zero, sice it is finite union of sets of measure zero, and so $\int_S f$ exist.
So could someone help me, please?