Measure Theory problem about rectangle cover.

Question

Show that if $A$ is a compact subset of $R^{n}$ and $A$ has measure zero in $R^{n}$, then given $\epsilon>0,$ there is a finite collection of rectangles of total volume less than $\epsilon$ covering $A$.

I know that countably finite collection of rectangle have total volume less than $\epsilon$ by definition of measure 0 set A. But how to prove finite collection of rectangle have measure 0.

Answer 1

Like you said, by definition of $A$ having measure $0$, given $\epsilon > 0$, there's a countable collection of open rectangles $U_1, U_2, U_3, \dots$ which cover $A$ and have total volume $< \epsilon$.

Now, $A$ is compact, and $\{U_j\}_{j=1}^{\infty}$ is an open cover of $A$; so by compactness ... (I leave it to you to complete the argument)

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Also, you're not trying to prove finitely many rectangles have measure $0$; in fact a non-trivial rectangle never has measure zero. You want to show the existence of a finite collection of rectangles with total volume $< \epsilon$.

Answer 2

There is an open set $U$ such that $A \subseteq U$ and $m(U) <\epsilon$. [$m(A)$ denotes the volume of $A$]. We can write $U$ as the union of sequence of open rectangles $R_n$. $A$ is covered by a finite number of these rectangles and their total volume is less than $m(U) <\epsilon$.