Some questions about Riemann integration

Question

I just learn some basic definition about

Let $f, g:[0,1] \rightarrow \Bbb R$ be

$ f(x) = \begin{cases} 1, & \text{if $x \in \Bbb Q$} \\ 0, & \text{otherwise} \end{cases}$

$g(x) = \begin{cases} 1, & \text{if $x=1/n, n=1,2,...$} \\ 0, & \text{otherwise} \end{cases}$

We know $f$ is not Riemann integrable, but $g$ is.

So my first question is, is it true that if the set of discontinuous points is a dense set, then that function is not Riemann integrable.

My second question is we know $h:[0,1] \rightarrow \Bbb R$ by $h(x)=1$ is integrable and has value $1$. So if we have a dense set $D$ in $[0,1]$ which cardinality of $D$ and $D^c$ are equal, and define $ u(x) = \begin{cases} 1, & \text{if $x \in D$} \\ 0, & \text{otherwise} \end{cases}$

Can we define a similar 'integral' to say the value of the 'integral' = $1/2$

Thank you!

Answer 1

The answer to the first question is no. Thomaes's function is discontinuous at the rational points and continuous at the irrationals, and Riemann integrable.

As for the second question, consider the following example. Let $a\in(0,1)$ and $f_a\colon[0,1]\to\Bbb R$ defined as $$ f_a(x)=\begin{cases} 1 & \text{if }x\in([0,a]\cap\Bbb Q)\bigcup([a,1]\setminus\Bbb Q),\\ 0 & \text{if }x\in([0,a]\setminus\Bbb Q)\bigcup([a,1]\cap\Bbb Q). \end{cases}$$ Any reasonable measure would give $\int_0^1f_a=1-a$, and not $1/2$.