Upper and Lower integrals of a piecewise function

Question

Let Q =I x I, where I = [0,1]. Define f:Q$\rightarrow$ R by letting f(x,y) = $\frac{1}{q}$ if y is rational and x= $\frac{p}{q}$, where p and q are positive integers with no common factor, let f(x,y) = 0 otherwise. Compute the upper and lower integral of $\int_{y \in I} f(x,y)$ I have no intuition whatsoever on how to attempt this question. Please explain your procedure in particular, if say we divide x in x=p/q and x $\not=$ p/q, then for y rational we will get 1/q and y irrational we get 0, how does the integral then even exist for x=p/q. Please be as detailed as possible, I've been stuck on this for quite some time and I would really appreciate all the help I can get.

Answer 1

I am guessing that $\int_{y \in I} f(x,y)$ is what might be written also $\int_0^1 f(x,y)\,dy$. It is a function of $x$. We need to consider it according to whether $x$ is rational or not.

If $x$ is rational then it'll be $p/q$ for some $p, q$ with no common factor. Then $f(x,y)=1/q$ for $y$ rational and $f(x,y)=0$ for $y$ irrational. I am guessing that the "lower integral" is the sup for $$\sum_{k=0}^{n-1} (y_{k+1}-y_k) \inf_{y_k\le y \le y_{k+1}} f(x,y)$$ taken over partitions $0=y_0 < y_1 < \cdots <y_{n-1}<y_n = 1$ (cf. Baby Rudin 6.1). It's easy to see that this is zero because every interval $[y_k,y_{k+1}]$ contains an irrational $y$, and so $f(x,y)$ for that $y$ will be zero, and that will be the inf in the sum. Making a similar guess for the upper integral, it's inf for $$\sum_{k=0}^{n-1} (y_{k+1}-y_k) \sup_{y_k\le y \le y_{k+1}} f(x,y)$$ also taken over the same partitions. This value will be $1/q$ since every interval $[y_k,y_{k+1}]$ contains a rational $y$, $f(x,y)=1/q$ for that $y$, and so that will be the sup in the sum, and note $$\sum_{k=0}^{n-1} (y_{k+1}-y_k) = 1$$ because of how the partitions are set up.

For $x$ irrational, $f(x,y)=0$ for all $y$, and so both upper and lower integrals are zero.