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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 $[0.1]$.

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    $\begingroup$ Do you mean Riemann or Lebesgue integrable? What is your definition if integrable? $\endgroup$ – user608030 Nov 29 '18 at 16:16
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    $\begingroup$ 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 $1$. $\endgroup$ – user3482749 Nov 29 '18 at 16:18
  • $\begingroup$ It's Riemann intergrable. I have not learned Lebesgue integrable. $\endgroup$ – Maggie Nov 29 '18 at 16:21
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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!

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  • $\begingroup$ Thank you! I understand! $\endgroup$ – Maggie Nov 29 '18 at 16:39
  • $\begingroup$ Wait! What is the mean of "These sums are independent of the partition itself "??? $\endgroup$ – Maggie Nov 29 '18 at 16:49
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    $\begingroup$ 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 $x_j$. $\endgroup$ – Josh B. Dec 3 '18 at 20:07

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