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I have to show that $$f: [0,1]\times[0,1]\rightarrow \mathbb{R}$$ $$f(x,y) = \left\{\begin{matrix} 1, & x\in\mathbb{Q}\\ 2y, & x \in\mathbb{R}-\mathbb{Q} \end{matrix}\right. $$

that the iterated Integral exist and calculate its value. Beside that I have to show that this function on this domain is not Riemann integrable.

I´ve shown in a ways that the iterated Integral exist and its 1, but not sure if it´s right: $$ \int_0^1 f(x,y) \, \mathrm{d}y = \begin{cases} \int_0^1 1 \, \mathrm{d}y & \text{if } x \in \mathbb{Q} \\ \int_0^1 2y \, \mathrm{d}y & \text{if } x \in \mathbb{R} \setminus \mathbb{Q}\end{cases} = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 1 & \text{if } x \in \mathbb{R} \setminus \mathbb{Q}\end{cases}. $$

And don´t quite know how to show that the function is not Riemann integrable.

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  • $\begingroup$ $f$ is a function of two variables. What definition of "Riemann integrable" are you using for functions of two variables? $\endgroup$ Nov 26 '19 at 19:02
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    $\begingroup$ You've shown correctly that $\int_0^1 f(x,y)\; dy = 1$. So the iterated integral $$\int_0^1 \int_0^1 f(x,y)\; dy \; dx = \int_0^1 1 \; dx = \ldots$$ On the other hand, there's another iterated integral $$ \int_0^1 \int_0^1 f(x,y)\; dx \; dy$$ That's where a question of Riemann integrability becomes important. $\endgroup$ Nov 26 '19 at 19:07
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Note that, with $y\neq 1/2$ fixed, $f(x,y) = 2y \neq 1$ for irrational $x$ and $f(x,y) = 1$ for rational $x$. Since the rationals are dense, the map $x \mapsto f(x,y)$ is everywhere discontinuous. Hence, $\int_0^1 f(x,y) \, dx$ fails to exist for almost every $y$ and the second iterated (Riemann) integral fails to exist.

Nonexistence of one iterated integral is not sufficient to prove that $\int_{[0,1]^2} f(x,y) \,d(x,y)$ fails to exist. A counterexample can be furnished upon request.

However, we can prove that $f$ is not Riemann integrable over $[0,1]^2$ directly by showing the infimum of upper Darboux sums (upper integral) does not equal the supremum of lower Darboux sums (lower integral).

Take any partition $P$ of $[0,1]^2$ with $m\cdot n$ subrectangles $R_{jk} =[x_{j-1},x_j]\times [y_{k-1},y_k]$ formed from points $0 = x_0 < x_1 < \ldots < x_m = 1$ and $0 = y_0 < y_1 < \ldots < y_n= 1.$

Let $q$ be the index such that $y_{q-1} < 1/2 \leqslant y_q$. Since each subrectangle contains points $(x,y)$ where $x$ is rational and where $x$ is irrational, we have, for all $1 \leqslant j \leqslant m$,

$$M_{jk} =\sup_{x \in R_{jk}}f(x,y) = \begin{cases}2y_k,& k \geqslant q \\1, & k < q \end{cases}\quad ,\,\,m_{jk} =\inf_{x \in R_{jk}}f(x,y) = \begin{cases}1,& k \geqslant q \\2y_k, & k < q \end{cases} $$

Noting that $M_{jk}$ is independent of $j$ and $\sum_{j=1}^m (x_j - x_{j-1}) = 1$, the upper sum is

$$\begin{align}U(P,f) &= \sum_{j=1}^m\sum_{k=1}^n M_{jk}(x_j- x_{j-1})(y_k - y_{k-1}) \\ &=\sum_{k=1}^{q-1} (1)(y_k - y_{k-1}) + \sum_{k=q}^{n}2y_k(y_k - y_{k-1})\\ &\geqslant y_{q-1} + \int_{y_{q-1}}^1 2y \, dy \\ &= y_{q-1} + 1 - y_{q-1}^2 \end{align} $$

We have $y_{q-1} \to 1/2$ as the partition norm $\|P \| \to 0$ and the upper Darboux integral must satisfy

$$\overline{\int}_{[0,1]^2} f(x,y) \, dy = \lim_{\|P\|\to 0}U(P,f) \geqslant 5/4$$

Similarly, we can show for lower sums that

$$L(P,f) \leqslant \int_{0}^{y_{q-1}} 2y \, dy + 1 - y_{q} = y_{q-1}^2 + 1 - y_{q-1},$$

and

$$ \underline{\int}_{[0,1]^2} f(x,y) \, d(x,y) = \lim_{\|P\|\to 0}L(P,f) \leqslant 3/4$$

Therefore, $f$ is not Riemann integrable over $[0,1]^2$ since the upper and lower Darboux integrals are not equal.

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