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 $

Solution 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 $m (\partial A) = m([0,1] \setminus A) > 0$.