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Let Let $f$ be a real valued map defined on $[0,1]$, given by $f(x)= 0$ if $x\in \mathbb{Q}$ and $x$ if $x\notin \mathbb{Q}$. Show that $f$ is not reimann integrable.

May I have hints on how to approach this problem, clearly, $sup_{D\in D[a,b]}$ $\{$ L(f,D) $\}$ $=$ $0$. Now, how do I show that $infU(f,D)$ $\neq 0$?

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  • $\begingroup$ Any function $g$ constant on an interval and having $g\ge f$ on that interval must be at least one on that interval, since intervals contain irrationals. That's the point, really – both the rationals and the irrationals are dense in the reals. $\endgroup$ – Gerry Myerson Oct 23 '19 at 0:25
  • $\begingroup$ HINT: Any interval (that's not just a point) contains both rational and irrational numbers. $\endgroup$ – Dzoooks Oct 23 '19 at 0:44
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Take any partition $P: 0 = x_0 < x_1 < \ldots < x_n = 1$.

Since there are irrational points in $[x_{j-1},x_j]$ arbitrarily close to $x_j$ we have $\sup_{x \in[x_{j-1},x_j]} f(x) = x_j$ and

$$U(P,f) = \sum_{j=1}^n x_j(x_j - x_{j-1}) > \sum_{j=1}^n \frac{1}{2}(x_j+ x_{j-1})(x_j - x_{j-1})= \frac{1}{2}\sum_{j=1}^n (x_j^2 - x_{j-1}^2) \\ = \frac{1}{2}(x_n^2 - x_0^2) = \frac{1}{2}$$

Hence, $\inf_P U(P,f) \geqslant \frac{1}{2} > 0$.

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  • $\begingroup$ On $[x_{j-1},x_j]$ we have $f(x) \leqslant x_j$ since for any $y \in [x_{j-1},x_j]$ either $f(y) = 0$ if $y$ is rational or $f(y) = y \leqslant x_j$ if $y$ is irrational. For any $\epsilon > 0$ with $x_j - \epsilon > x_{j-1}$ there exists an irrational $\xi \in [x_j-\epsilon,x_j)$ such that $f(\xi) = \xi > x_j-\epsilon$. Together these imply that the supremum of $f$ on this interval is $x_j$. $\endgroup$ – RRL Oct 23 '19 at 4:57

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