A bounded set $S\in\mathbb{R}^n$ is called rectifiable if $\int_S1$ exists (the integral here is Riemann integral.) Let $S$ be a bounded set in $\mathbb{R}^n$ that is the union of the countable collection of rectifiable sets $S_1,S_2,\cdots$. Show that $S_1\cup \cdots\cup S_n$ is rectifiable, and give an example showing that $S$ need not be rectifiable.
To show that $S_1\cup \cdots\cup S_n$ is rectifiable, it suffices to show that the union of two rectifiable sets is rectifiable. But this follows from the fact that if $f$ is integrable over $A$ and over $B$, then it is integrable over $A\cup B$.
For the second part, I'm thinking of something like $S\in\mathbb{R}$, $S_1=[0,1],S_2=[1,1+1/2],S_3=[2,2+1/3], \cdots$, but the problem here is that the set $S$ is not bounded. What could an example be?
Best Answer
Note that if for each $S_i$, $\displaystyle \int_{S_i}1$ exists, for each $\epsilon >0$ we can find a finite number of closed rectangles $\{R_{1i},\ldots,R_{k_ii}\}$ that cover $\partial S_i$ and whose volumes sum less than $\epsilon$. This is due to compactness of $\partial S_i$. In such a case, any finite union of such sets will have this property too. What about this? Enumarate the rationals in $[0,1]$, and draw lines from each of them to their pair in the line trough $y=1$. Then consider the set that is the union of all this lines. It should look like a "codebar" of rational endpoints. This is inside the unit square, but the Riemann integral cannot exist. Indeed, the boundary of this set is $[0,1]\times [0,1]$.