Let $B\subset \mathbb{R}^n$ be a closed rectangle, $f:B\to\mathbb{R}$ Reimann integrable and $A\subset B$ a set of points where $f$ is not continous.
I`m trying to show that the measure of $\partial A$ is $0$ ,where $\partial A$ is the boundary of $A$.
I know that the measure of $A$ is $0$,and that for general $0$ measure $A$ the statement isn`t correct, However, when $A$ is defined as the set of discontinuity points of an integrable function, I can't find a counter example.
My attempt of a proof was:
Take $A_\epsilon=\{x\in B : osc(f,x)\geq\epsilon\}$
,and we have $A=\bigcup_{k=1}^\infty A_{\frac{1}{k}}$
We have $\partial A\subset \bigcup_{k=1}^\infty \partial A_{\frac{1}{k}}$ so it is sufficient to show $\partial A_{\frac{1}{k}}$ is of measure zero.
$A_{\frac{1}{k}}$ is compact, and of measure $0$ (Following a proof I saw of the Lebesgue theorem), so we can find a finite cover of rectangles as small as we want, this is also a cover of the interior of $A_{\frac{1}{k}}$, so the interior is of measure 0. Now we have $int(A_{\frac{1}{k}})\cup\partial A_{\frac{1} {k}}=A_{\frac{1}{k}}$ ($A_{\frac{1}{k}}$ is closed).
(*)Since the measure of $int(A_{\frac{1}{k}})=0$ and also the measure of $A_{\frac{1}{k}}=0$, we must have $\partial A_{\frac{1}{k}}=0$
I`m unsure of my proof since because honestly, I find it hard to believe the statement is true.
I`d be glad if someone can provide a counter example, or verify me and tell me this statement is actually correct.
Best Answer
For a counterexample, take $n = 1$, $B = [0,1]$, $A = [0,1] \cap \mathbb{Q}$ and
$$f(x) = \begin{cases}\frac{1}{q}, & x \in A, \,\,x = \frac{p}{q} \text{ in lowest terms}\\0, & x \in [0,1] \setminus A \end{cases}$$
This is Thomae's function which is integrable on $[0,1]$.
Note that $f$ is discontinuous at every point in $A' =A \setminus \{0\}$ -- the rationals in $(0,1]$
However, $\partial A' = [0,1]$ and $m (\partial A') = 1 \neq 0$.