# Function and Integrability

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 $$infU(f,D)$$ $$\neq 0$$?

• 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. – Gerry Myerson Oct 23 '19 at 0:25
• HINT: Any interval (that's not just a point) contains both rational and irrational numbers. – Dzoooks Oct 23 '19 at 0:44

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, $$\inf_P U(P,f) \geqslant \frac{1}{2} > 0$$.
• 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 $x_j$. – RRL Oct 23 '19 at 4:57