Show that a double integral of $f$ exists on the closed unit square

Question

Suppose $f$ is defined in the square $S=\{(x,y) | 0 \leq x \leq 1, 0 \leq y \leq 1\}$ by \begin{cases} 1 &\quad\text{if x is irrational}\\ 4y^3 &\quad\text{if x is rational}\\ \end{cases}

a) Show $\int_0^1 \! [\int_0^1 \! f(x,y) \, \mathrm{d}y ] \, \mathrm{d}x$ exists and is equal to $1$

b) Show that $\int_S \! f(x,y) \, \mathrm{d}V$ does not exist.

This is the generalized Riemann integral. I know that f takes on the values $1$ and $4y^3$ within every interval in every subdivision of $S$. Honestly, that's about all I know.

Answer 1

The first part follows by evaluating the inner integral for $x$ rational and irrational separately which yields 1. This is integrable on $[0,1]$.

For the second, take any rectangle $R=[x_1,x_2]\times [y_1,y_2] \subset [0,1]^2$.

It is clear that $\sup_{(x,y) \in R} f(x,y) = \max(1, 4y_2^3)$ and $\inf_{(x,y) \in R} f(x,y) = \min(1, 4y_1^3)$.

Note that the functions $\phi_1(x,y) = \max(1,4y^3)$, $\phi_2(x,y) = \min(1,4 y^3)$ are integrable on $[0,1]^2$ since they are continuous. It is easy to see that $\int \phi_1 > 1 > \int \phi_2$.

From the above, we see that for any partition $P$ of $[0,1]^2$ we have $U(f,P) = U(\phi_1,P)$, $L(f,P) = L(\phi_2,P)$, so it follows that $\inf_P U(f,P) = \inf_P U(\phi_1,P) = \inf \phi_1 > 1$ and $\sup_P L(f,P) = \sup_P U(\phi_2,P) = \sup \phi_2 < 1$, and so $f$ is not integrable.

Answer 2

For the first one, you are holding x fixed and evaluating it as a single function of y. Note that if x is rational, then $\int _0 ^1 f(x,y)dy=\int _0 ^1 1\cdot dy=1$. Likewise if x is irrational, we have $\int _0 ^1 f(x,y)dy=\int _0 ^1 4y^3dy=1.$ So, the inner integral evaluates to be 1 as we hold x fixed. Thus the nested integral simplifies to $\int _0 ^1 dx=1$

However, if we don't do it nested, then you can show the set of discontinuities of $f$ is of measure greater than 0, thus is not riemann integrable. (Equivalently, you can show that you can never get a lower sum and upper sum to be within a given $\epsilon$ due to density as you noted)

Answer 3

The first part is neatly addressed by another answer. For the second part, show that if you take a uniform subdivision into $n^2$ smaller rectangular squares of side length $1/n$, then when $y$ is close enough to $0$ the upper integral and lower integral obtained from the subdivision are not equal, and in fact are different by at least some constant amount.