For Lebesgue integrability, for all subsets $A$ of $[0,1]$, the indicator function $1_A$ is Lebesgue integrable iff $A$ is measurable.
Is there any similar characterization of Riemann integrability for indicator functions in terms of denseness? For example, it is well-known that:
- if $A\subseteq[0,1]$ and $[0,1]\setminus A$ are both dense in $[0,1]$, then $1_A$ is not integrable;
- $1_A$ is Riemann integrable iff it is continuous almost everywhere.
I initially guessed that for all subsets $A$ of $[0,1]$, the indicator function $1_A$ is not Riemann integrable iff $A$ and $[0,1]\setminus A$ are both dense in $[0,1]$, but that's not true -- just take something like $A=\mathbb{Q}\cap[1/4,3/4]$, and then there are no elements of $A$ in the interval $(0,1/4)$, so $A$ isn't dense in $[0,1]$. My conjecture now is that function $1_A$ is not Riemann integrable iff $\exists\varnothing\neq B\subseteq[0,1]$ s.t. $A$ and $B\setminus A$ are both dense in $B$.