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Let $f:[0,1]\rightarrow\mathbb{R}$ be a function which has the value $$f(x)=\left\{ \begin{array}{l l} x^2 & \quad ,x\in\mathbb{Q}\\ x^3 & \quad ,x\notin\mathbb{Q} \end{array} \right.$$ Examine the integrability of $f$ in $[0,1]$.

So far, I have considered a random partition $P$ of $[0,1]$ and I have tried to calculate $|U-L|$ but with no result.

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  • $\begingroup$ Your function is not continuous except at $0$ and $1$, so you should expect it to not be integrable. What happens when you deal with the upper and lower sums? (Hint: for $x \in (0,1)$, which is bigger, $x^2$ or $x^3$? How much bigger is it?) $\endgroup$ – Ian Feb 25 '15 at 23:50
  • $\begingroup$ @Ian I can see that $x^3<x^2$ but later when I try to calculate the upper and lower sums I can't figure out which $\inf(f(x))$ and $\sup(f(x))$ for $x\in(x_k,x_{k+1})$ to use. Or at least how to bound them. $\endgroup$ – chrispap Feb 25 '15 at 23:58
  • $\begingroup$ Since $x^3$ is smaller and both the rationals and the irrationals are dense, your infimum will be $x^3_k$ and your supremum will be $x^2_{k+1}$. $\endgroup$ – Ian Feb 26 '15 at 0:15
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There is a theorem which states that a function is integrable if and only if the set of discontinuities for the function has "measure" zero, i.e. is "negligible". But the "measure" of $(0,1)$ is $1$ and your function is discontinuous everywhere there. So according to the theorem, your function is not integrable. If you want a more basic proof, just break up your interval $[0,1]$ into $n$ subintervals of equal length in $[0,1]$ and then calculate the supremum and infimum of your function on each interval so that you get the upper and lower integrals, and show that they are not equal in the limit. (Hint: You should get that the upper integral in the limit is $\int_0^1 x^2 dx$ and the lower integral in the limit is $\int_0^1 x^3 dx$.) You don't have to brute force calculation to do this, just use the definition of integral and properties of rationals and irrationals.

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