Give an example where $\int_{\bar A} f$ exists but $\int_A f$ does not for a continuos $f$ on a bounded open subset $A$ of $\mathbb{R}^n $

Question

In the book of Analysis on Manifolds by Munkress, at page 121, it is asked that

Let $A$ be a bounded open set in $\mathbb{R}^n $; let $f: \mathbb{R}^n > \to \mathbb{R} $ be a bounded continuous function. Give an example where $\int_{\bar A} f$ exists but $\int_A f$ does not.

I couldn't find any such example. I mean I do noticed that $A$ should not be rectifiable, and the only way I can think of that $f$ is not integrable on $A$ is that the set $E$ where the limit $$\lim_{x \to x_0} f(x) = 0$$ fails for $x_0 \in \partial A$ has a non-zero measure, but then this set $E$ also contained in $\bar A$, so in such a case $\int_{\bar A} f$ also wouldn't exists.

In short, I'm looking for such an example.

Note: I have read this answer, but I have no idea what the answerer is talking about, so I wanted to ask a new question including my own thoughts.

Answer 1

The key here is that the Riemann integral of a bounded function $f:A \to \mathbb{R}$, where $A \subset \mathbb{R}^n$ is bounded but not a closed rectangle, is defined as

$$\int_A f = \int_Q f \chi_A,$$

where $Q$ is any closed rectangle containing $A$.

Take $n=1$, $A$ as the set of rational numbers in $[0,1]$ and $f: x \mapsto f(x) = 1$. Since $\bar{A} = [0,1]$ we have existence of the integral $\int_{\bar{A}}f = \int_0^1f(x) \,dx =1$.

However, the boundary $\partial A$ of $A$ is the set of irrational numbers in $[0,1]$ and is not of measure zero. Hence, $\int_A f = \int_{\bar{A}} f \chi_A$ does not exist, since $f\chi_A = \chi_A$ is discontinuous on the boundary $\partial A$.

Edit

I overlooked the requirement that $A$ is an open set. However, as pointed out by Daniel Fischer the same argument applies if $A$ is taken as the complement of a fat Cantor set in $[0,1]$. In this case $A$ is open with $m (\partial A) = m([0,1] \setminus A) > 0$.

Answer 2

This problem is Exercise 6 on p.121 in "Analysis on Manifolds" by James R. Munkres.

Example 1 on p.113:
We construct a bounded open set $A$ in $\mathbb{R}$ such that $\operatorname{Bd}A$ does not have measure zero.

The rational numbers in the open interval $(0,1)$ are countable; let us arrange them in a sequence $q_1,q_2,\dots$. Let $0<a<1$ be fixed. For each $i$, choose an open interval $(a_i,b_i)$ of length less than $\frac{a}{2^i}$ that contains $q_i$ and is contained in $(0,1)$. These intervals will overlap, of course, but that doesn't matter. Let $A$ be the following open set of $\mathbb{R}$: $$A=(a_1,b_1)\cup (a_2,b_2)\cup\dots.$$
$\dots$
Because $A$ is a subset of $[0,1]$ that contains each rational in $(0,1)$, we have $\overline{A}=[0,1]$.

Then, the author proved that $\operatorname{Bd}A$ doesn't have measure zero.

So, $A$ is not rectifiable, but $\overline{A}=[0,1]$ is rectifiable.
So, $\int_{\overline{A}}1$ exists but $\int_A 1$ does not.