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

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 $$[0.1]$$.

• Do you mean Riemann or Lebesgue integrable? What is your definition if integrable? – user608030 Nov 29 '18 at 16:16
• What form of integral are you using? Presumably not Lebesgue, because that function is Lebesgue integrable on $[0,1]$ (with integral $1$). Riemann or Regulated, then? It's not regulated because it's not a uniform limit of step functions. It's not Riemann integrable because there are sequences of tagged partitions whose Riemann sums converge to both $0$ and $1$. – user3482749 Nov 29 '18 at 16:18
• It's Riemann intergrable. I have not learned Lebesgue integrable. – Maggie Nov 29 '18 at 16:21

Let $$0=x_0 be any partition of the interval $$[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.
• What I mean is that regardless of the partition used, the upper sum is $1$ and the lower sum is $0$; in other words, the fact that we got $1$ and $0$ is unrelated to any of the partition points $x_j$. – Josh B. Dec 3 '18 at 20:07