Riemann integrability on open subset of $\mathbb{R}$

Question

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function such that $f$ is Riemann integrable over $\mathbb{R}$.

Is it true that $f$ is Riemann integrable over any open subset $A\subset \mathbb{R}$??

Answer 1

Assuming you mean Riemann integrable on any closed, bounded interval and you have defined the Riemann integral over an arbitrary open set using the characteristic function, the answer is no.

We can construct an open set $A \subset [0,1]$ whose boundary does not have measure zero. Take $f(x) = 1$.

Then the Riemann integral

$$\int_A f := \int_0^1 f(x) \chi_A(x) \, dx $$

does not exist

Construction of $A$

Let $\{r_i\}$ be an enumeration of the rational numbers in $(0,1)$. This is a countable set. We can take intervals such that $r_i \in(a_i,b_i) \subset (0,1)$ and $\sum_{i \geqslant1} (b_i - a_i) < 1$. The union

$$A = \bigcup_{i \geqslant 1}(a_i,b_i)$$

is an open set with boundary $\partial A= [0,1] \setminus A$ since the rationals are dense in $[0,1]$. However, $m(\partial A) = m([0,1]) - m(A) > 1 - \sum_{i \geqslant 1}(b_i-a_i)> 0$.