I have $Q=[0,1]\times[0,1]$ and $f:Q\to \mathbb R$ defined by
$$f(x,y)= \begin{cases} 1, & \text{if } x \ne {1\over 2} ; y \in [0,1] \\ 1, & \text{if } x = {1 \over 2} ; y \in [0,1]\setminus {\mathbb Q} \\ 0, & \text{if } x ={1\over 2} ; y \in [0,1]\cap {\mathbb Q}. \end{cases}$$
The function $f$ is Riemann-integrable on $Q$ but the $1 \over 2$-section isn't integrable. How can I prove it?
On
For small $\epsilon > 0,$ partition $Q$ into $[0,1/2-\epsilon]\times [0,1], [1/2-\epsilon, 1/2+\epsilon]\times [0,1], [1/2+\epsilon]\times [0,1].$ For this partition, call it $P,$ we have
$$ 1\cdot (1/2-\epsilon)\cdot 1 + 0\cdot 2\epsilon \cdot 1 +1 \cdot (1/2-\epsilon)\cdot 1 \le L(P,f).$$
So we have $1-2\epsilon \le L(P,f) \le U(P,f)\le 1.$ Thus the difference between upper and lower sums of $f$ is arbitrarily small, which implies $f$ is Riemann integrable on $[0,1]^2.$
Partition $Q$ into $n^2$-many squares each of area $\frac1{n^2}$. (Use a grid of vertical and horizontal lines $x=\frac in$, $y=\frac jn$, $0\le i,j\le 1$.) Exactly $2n$-many of these squares touch the vertical line $x=\frac12$. In these $2n$-many squares you have a choice to pick the sample points $x^*$ so that either $f(x^*)=0$ or $f(x^*)=1$. In all other $(n^2-2n)$-many squares we have $f(x^*)=1$ regardless of the choice of sample points. Therefore each Riemann sum $R_{n^2}$ corresponding to such a partition evaluates to at least $(n^2-2n)\cdot\frac1{n^2}=1-\frac2n$ and at most $n^2\cdot\frac1{n^2}=1$, i.e. $1-\frac2n\le R_{n^2}\le 1$. Since $1-\frac2n\to1$, by the squeeze theorem we have that $R_{n^2}\to1$, that is the Riemann integral exists and equals $1$.
If we restrict our attention only to the section $x=\frac12$, then in each vertical interval of the form $\{\frac12\}\times[\frac{j-1}n,\frac jn]$ we could pick sample points $x^*$ at which $f(x^*)=1$ as well as other sample points $x^*$ at which $f(x^*)=0$. The vertical line segment $\{\frac12\}\times[0,1]$ is partitioned into $n$-many vertical line segments $\{\frac12\}\times[\frac{j-1}n,\frac jn]$ each of length $\frac1n$ (for $j=1,\dots,n$), hence if we use sample points $x^*$ with $f(x^*)=1$, then the corresponding Riemann sum $R_n=n\cdot\frac1n=1$. If we use sample points $x^*$ with $f(x^*)=0$, then the corresponding Riemann sum $R_n=n\cdot\frac1n\cdot0=0$. Depending on the choice of sample points we could have $\lim_n R_n=1$ or $\lim_n R_n=0$. The integral does not exist, since in order to exist the limit of the $R_n$ should exist and be independent of the choice of sample points.