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Let $S$ be a bounded set in $\mathbb{R}^n$ that is the union of the countable collection of rectifiable sets $S_1, S_2, \dots$.

How can we show that $S_1 \bigcup \dots \bigcup S_n$ is rectifiable? Also, is there an example showing that S need not be rectifiable?

My proof:

From the definition of rectifiable, if we look at a bounded set S in R^n, then whose volume over S of the constant function 1 is integrable.

Also, A subset S of R^n is rectifiable iff S is bounded and Bd S has measure Zero.

So, we are told that S is bounded and we need to show that Bd(S) is zero. For S is bounded, and Bd(S) = Reals, which has measure zero.

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Probably you need to add a definition (or reference) for "rectifiable set". –  GEdgar Feb 14 '12 at 12:32
    
pls see above.. –  James R. Feb 16 '12 at 10:46

1 Answer 1

With the definition, we can see your answer. You need to prove: $\text{Bd}(A \cup B) \subseteq \text{Bd}A \cup \text{Bd}B$. Then it follows that a finite union of rectifiable sets is rectifiable.

I assume when you write $S_1 \bigcup \dots \bigcup S_n$ you mean a finite union. Plus I assume it was a mistake that the number $n$ of sets in the union is the same as the dimension of your space $\mathbb R^n$.

On the other hand, your statement has "countable" in there, too. Did you want that? A countable union of rectifiable sets need not be rectifiable. For example, a singleton $\{r\}$ is rectifiable, but a countable set, dense in the unit ball, is not rectifiable, although it is a countable union of those singletons.

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