On the domain $D:[0,1]\times[0,1]\times[0,1], ~f(x,y,z)$ is defined to be: $$ \begin{align*} &1~ \text{for rational}~ x,\\ &2~ \text{for irrational}~ x,~0\leq y \leq \frac{1}{2}\\ &0~ \text{for irrational}~ x,~\frac{1}{2}< y \leq 1. \end{align*} $$

The question asks to show that the iterated integral $\int_0^1\int_0^1\int_0^1f(x,y,z)~dz~dy~dx$ exists, but $f$ is not integrable.

I assume that the value for the iterated integral is simply 1, since $\mathbb{Q}$ is countable but $\mathbb{R}\setminus\mathbb{Q}$ is uncountable, and thus it is safe to assume that $x$ will take on irrational values for most of the interval $[0,1]$, with rational values negligible.

However, would that be a precise answer?

Also, I understand that $f$ would be nowhere continuous, but I am not sure how to extend that thought to rigorously show that $f$ would not be integrable.

My attempt was to show that an iterated Riemann sum would take on the form of: $$\sum_i^n \frac{1}{n}~ f(x_i,y_i,z_i),$$ but since there are infinitely many both irrational and rational $x$ in any given interval, $f(x_i,y_i,z_i)-f(x_{i+1},y_{i+1},z_{i+1})$ would always yield a nonzero value, which by definition should tend to $0$ should the number of iterations approach infinity.

Thanks for any help!


1 Answer 1


It might be more illuminating if you show that you obtain different limits for different choices of the sample points.

For example (A). Pick $x$ always to be rational. Then each sum is $1$, the limit is $1$.

(B) Pick rational $x$ for $0\le y\le \frac12$ but irrational $x$ for $\frac12<y\le1$. Then each sum evaluates to $\frac12$ and the limit is $\frac12$.

(C) Pick irrational $x$ for $0\le y\le \frac12$ but rational $x$ for $\frac12<y\le1$. Then each sum evaluates to $\frac32$ and the limit is $\frac32$.

Since the limit is not supposed to depend in the choice of the sample points, the limit (and the Riemann integral) does not exist.


You must log in to answer this question.