1
$\begingroup$

A set of measure zero in $\mathbb{R^n}$ is such that for each $a>0$, it can covered by countably many open boxes (hyperrectangles), the sum of the volumes of these boxes being less than $a.$

How can I prove that we can replace open hyperrectangles by open hypercubes or open balls?

I tried to cover a rectangle by countable cubes in desired way then cover cubes by countable balls.

I can cover cubes by balls but I think first one is so hard!

$\endgroup$
0
$\begingroup$

Let's work with $n=2$ for simplicity - hopefully the general picture will become clear.

Suppose I have an open rectangle $R$ with dimensions $l\times w$, and $\epsilon>0$; I want to cover $R$ with open squares $S_1, . . . , S_n$ such that the "excess" $(\bigcup_{1\le i\le n} S_i)\setminus R$ has area $<\epsilon$.

The goal is to show that this can always be done - then the two notions of measure zero are the same, since given a "small-area" cover of a set by rectangles, I can just cover the rectangles by squares in a "reall-small-excess" way, and that gives a "small-area" cover of the original set by squares.

Rather than do this directly, I'll describe how to cover $R$ with closed squares with appropriately small excess. We can then expand each square a tiny amount to get a covering by open squares.

For positive $a$, let $l_a$ be the unique number in $[0, a)$ such that for some integer $k_l$, we have $k_la=l_a+l$. Intuitively, $l_a$ is the excess I get from covering a line of length $l$ by lines of length $a$. Similarly, let $w_a$ be the unique number in $[0, a)$ such that for some integer $k_w$, we have $k_wa=w_a+w$.

Well, now imagine covering $R$ by closed squares with dimensions $a\times a$, starting in the "bottom-left"; this winds up using $k_l$ squares lengthwise, and $k_w$ squares widthwise, for a total of $k_lk_w$-many squares. The total area of these squares is $k_lk_wa^2$, but we don't care about that; we care about the excess area $k_lk_wa^2-lw$. This excess is the area of a "strip" on the top and right of $R$ - specifically, the excess has area $$E_a=l_a(w+w_a)+w_a(l+l_a)-w_al_a$$ (draw it to see why). But then $$E_a<l_a(w+w_a)+w_a(l+l_a)<a(w+a)+a(l+a)=a(w+l+2a);$$ if we pick $a\le 1$ (say), this gives $E_a<a(w+l+2)$. But this lets us control $E_a$: just take $$a=\min\{1, {\epsilon\over w+l+2}\}.$$ Then if we cover $R$ by small squares of dimension $a\times a$, as described above, we'll only exceed the area of $R$ by $<\epsilon$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.