The following question is asked in the book of Analysis On Manifolds by Munkres, and given at page 103 as question 3.

$Q=A\times B$,where $A$ is a rectangle in $\mathbb{R^k}$ and $B$ is a rectangle in $\mathbb{R^n}$.

  1. Give an example where $\int_{Q}f$ exists and one of the iterated integral

$$\int_{x\in A}\int_{y\in B} f(x,y) \; \text{and} \; \int_{y\in B}\int_{x\in A} f(x,y)$$ exists, but the other does not.

  1. Find an example where both the iterated integrals of 1. exist, but the integral $\int_Q f$ does not. [Hint: One approach is to find a subset $S$ of $Q$ whose closure equals $Q$, such that $S$ contains at most one point on each vertical line and at most one point on each horizontal line.]

I'm having difficulty coming up with examples for the second case using the hint.

For the first one, I found $f(x,y)= 1/n$ if $y$ is rational and $x=m/n$, where $(m,n)=1$.

However, I have no clue about the second one, and what the hint even means.

I'd appreciate it if anyone can help me with this problem.

  • $\begingroup$ It would be nice to see what was the author thinking originally, or an answer to the second part is given in math.stackexchange.com/a/393272/279869 $\endgroup$
    – Our
    Jul 13, 2018 at 9:23
  • $\begingroup$ @onurcanbektas: See posted answer which discusses the linked answer you provided. $\endgroup$
    – RRL
    Jul 14, 2018 at 19:29
  • $\begingroup$ @RRL Thanks a lot for letting me know. $\endgroup$
    – Our
    Jul 14, 2018 at 19:45

1 Answer 1


Suppose $S$ is a countable, dense subset of $[0,1]^2$ and $f = \chi_S$. Here $f$ is not Riemann integrable on $[0,1]^2$ since, given any partition $P$, each rectangle contains points where $f(x) = 1$ and $f(x) = 0$ and the upper Darboux sum $U(P,f) = 1$ while $L(P,f) = 0$.

On the other hand, both iterated integrals equal $0$. For any $x \in[0,1]$ there is only one point $(x,y_x)$ such that $f(x,y_x) = 1$, but $f(x,y) = 0$ if $y \neq y_x$. Thus,

$$\int_0^1 f(x,y) \, dy = 0 \implies \int_0^1 \left(\int_0^1 f(x,y) \, dy \right)\, dx = 0,$$

with a similar argument and result for the other iterated integral.

It is not too difficult to construct such a set $S$ starting with any countable, dense subset $\{(x_j,y_j): j \in \mathbb{N}$} and then choosing

$$(\hat{x}_1,\hat{y}_1) = (x_1,y_1), \\ (\hat{x}_2,\hat{y}_2) \in B((x_1,y_1), 1/2) \text{ with }(\hat{x}_2,\hat{y}_2) \neq (x_1,y_1), \\ (\hat{x}_3,\hat{y}_3) \in B((x_3,y_3), 1/3) \text{ with }(\hat{x}_3,\hat{y}_3) \neq (x_1,y_1), (x_2,y_2), \\ \ldots $$

The linked answer in the comment provided by @onurcanbektas uses the function

$$f(x,y) = \frac{x^2 - y^2}{(x^2+y^2)^2}$$

This serves better as an example where the conclusion of Fubini's theorem is false for Lebesgue integrals because $|f|$ is not integrable. In the context of Munkres' question, $f$ in this case fails to be Riemann integrable simply because it is unbounded. (The hint leads to an example where the function is bounded.) Another technicality arises in consideration of the unequal iterated integrals as Riemann integrals. For example,

$$\int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dy = \begin{cases} \frac{1}{1+x^2}, & x \neq 0 \\ -\infty, & x = 0 \end{cases}$$

where $\int_0^1f(0,y)\, dy$ does not exist as a Riemann integral. Of course, we can bypass this issue by recognizing that the non-existence occurs on a set of measure zero and either accepting $\pm \infty$ as extended real numbers or reassigning the value.


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