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!
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 $<\epsilon$.