Show that $f$ is not Riemann integrable on $[0,1]$

Question

If $x$ is any rational number, $f(x)=0$.

If $x$ is any irrational number, $f(x)=1$.

I know that $f(x)$ oscillate between $0$ and $1$ on $[0.1]$. But I have not idea why it isn't integrable on $[0.1]$.

Answer 1

I assume that you mean that this function is not Riemann Integrable, as this function is actually Lebesgue Integrable. A function is defined as Riemann integrable if the upper sums and the lower sums of arbitrary partitions get arbitrarily close.

Let $0=x_0<x_1<...<x_{n-1}<x_n=1$ be any partition of the interval $[0,1]$. On any interval from $[x_{j-1},x_j]$, there exists a rational number and an irrational number on this interval. As such, the upper estimate on this subinterval is $1$ (since there is an irrational number on this interval) and the lower estimate is $0$ (since there is a rational number). Thus, the complete upper sum is $1$ and the complete lower sum is $0$. These sums are independent of the partition itself, so they can never get any closer, and the function is not integrable.

Hope this helps!