Dirchlet Riemann Integrable in certain interval

Question

Considering the Dirichlet function

f : f (x) = { 1 if x is rational 0 if x is irrational }

I want to know if this function can be Riemann Integrable in the interval [0,1].

I reasearched a bit, and get a bit confused, becouse in wikipedia says is not Riemann integrable the function,but saw in some web that says that in Riemann integrable with integral zero on interval [0,1] , so in conclusion, the function is Riemann Integrable on the interval [0,1]?

Thanks

Answer 1

The function is not Riemann-integrable because it is nowhere-continuous, and a function $f$ is Riemann-integrable only if the set of discontinuities of $f$ has measure zero and $f$ is bounded:

To show $f$ is nowhere-continuous, consider first a Rational number $x$. Then, since Irrationals are dense in Reals, every ball $B(x,e); e>0$ about $x$; no matter how small, will contain some Irrational, so that $f$ will contain values $0$ and $max|f(x)-f(y)|=1$ in $B(x,e)$, for any $e>0$ . A similar argument holds for $y$ Irrational, since the Rationals are also dense on the Reals, so that any ball $B(y,e)$ ; $e>0$ will contain an Irrational, so that $f$ will oscillate between $1$ and $0$ in $B(x,e)$, for any $e>0$.

Edit: Paraphrasing Pedro's answer below, the Lebesgue integral of $f$ is given by:

$1$m{$(x: f(x)=1$}+$0$m{$x:f(x)=0$}=$0+0=0$, since $m(\mathbb Q)=0$ , where $m$ is the Lebesgue measure , so that the Lebesgue sum converges.

Answer 2

The function is not Riemann integrable for the upper and lower integrals do not coincide. The lower integral is $0$, but the upper integral is $1$. Can you prove this? You need to use the density of both rationals and irrationals in $\Bbb R$.

On the other hand, the function is Lebesgue integrable and its integral is zero, since $f=0$ almost everywhere.