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I'm learning about Riemann integrable function and need help with the following problem:

Consider the function

$$f(x) = \begin{cases} x, & 0 \leq x \leq 1 \, \text{and} \, x \in \mathbb Q; \\ x^2, & 0 \leq x \leq 1 \, \text{and} \, x \notin \mathbb Q. \end{cases}$$

Examine if $f$ is Riemann integrable.

The definition the author is using for Riemann integrable function is the one comparing the upper/lower sums, i.e. a function $f$ is Riemann integrable if and only if

$$\inf_P\{U(f, P)\} = \sup_P\{L(f, P)\}.$$

This exercise follows from the classic example of the Dirichlet function which is not Riemann integrable on any closed interval. I was able to follow and understand the latter example very well by I'm having difficulties applying a similar method of resolution for this newer function. My intuition tells me that the considered function is not Riemann integrable given that $f(x)$ is analogous to $\chi_{\mathbb Q}$, the Dirichlet function.

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Consider $L(f,P)$ for an arbitrary partition $P$ and note that $x^2 \leq x$ in the interval and hence when considering the lower sums, we only need to worry about $x \in \mathbb{R} \setminus \mathbb{Q} $. In any given subinterval $[t_{i-1}, t_i]$ of the partition, the $\inf$ of $f$ is precisely equal to the $\inf$ of $g:[t_{i-1}, t_i] \to \mathbb{R}$ given by $g(x) = x^2$. The reason is because $g$ is continuous and increasing and there exist irrationals arbitrarily close to $t_{i-1}$. It thus follows that the $\sup$ of the lower sums is equal to $\int_0^1 x^2 \ dx = \frac 13$.

Analogously, we can show that the $\inf$ of the upper sums is $\int_0^1 x \ dx = \frac 12$.

The function is thus not Riemann integrable.

Alternatively, simply note that $f$ is discontinuous everywhere in $(0,1)$ and use Lebesgue's criterion.

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  • $\begingroup$ Thank you for your detailed and clear explanation. I have accepted your answer but I was wondering if you could tell me a little more about the Lebesgue criterion you are mentioning at the end of your answer. I have a basic knowledge in measure theory but I don't know which criterion you are referring to. $\endgroup$
    – glpsx
    Jun 2 '16 at 8:32
  • $\begingroup$ @VonKar This is the Lebesgue Criterion for Riemann integrability. Basically, a function is Riemann integrable if and only if it is bounded and continuous almost everywhere (formally, its set of discontinuities has measure $0$). This function is discontinuous on $(0,1)$ so it cannot be Riemann integrable. $\endgroup$
    – AlohaSine
    Jun 2 '16 at 14:23

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