About equivalent measure on $\mathbb{R}^n$ Let $m$ denote the canonical Lebesgue measure on $\mathbb{R}^n$ and $\mu$ be a regular Borel measure on $\mathbb{R}^n$, if for all open balls $B \subset \mathbb{R}$, we have
$$C_1\ m(B)\leq \mu(B) \leq C_2 m(B)$$
for some positive $C_1>0,C_2>0$. The problems asks to show $\mu,m$ is mutually absolutely continuous.
The conclusion will following immediately if we can show
$$C_1\ m(E)\leq \mu(E) \leq C_2 m(E)$$
for all Borel set $E$. By regularity it shall be sufficient to reduce this to the case for open set $E$. If we are dealing with dimension 1, then any open set can be decomposed as countable disjoint open intervals and we are done. But for $n\geq 2$, open set does not generally admit such decomposition(i.e. the union of disjoint open balls). Though it is obviously that open sets can be covered with open balls within any given error in measure, it is hard to expect we can take some cover of disjoint balls, so the argument still does not hold. I am not sure if this approach to the proof works, any comment shall be greatly appreciated!
[Editted]
Thanks to the comment below I have reached conclusion by another approach. Now I have another question: If we let $f:=d\mu / dm$, then intuitively $f$ should be bounded both from below and above, i.e. $f,1/f \in L^\infty(m)$, does this hold true?
 A: To prove that $f$ and $1/f$ are in $L^{\infty}$: if $E$ is a Borel set, then $$C_1m(E)\leq\mu(E)=\int_Efdm\Rightarrow\int_E(f-C_1)dm\geq 0.$$ Since this is true for all Borel subsets of $\mathbb R^n$, you have that $f\geq C_1$ almost everywhere. Similarly, $f\leq C_2$ almost everywhere.
A: I submit a possible answer based on the heuristic principle "boxes are the same as balls", which I have sometimes heard at seminars in real analysis. 
Let $E\subset \mathbb{R}^n$ be open. We can decompose $E$ into a countable and disjoint union of dyadic boxes: 
$$E=\bigcup_{j=0}^\infty D_j.$$
Therefore we only need to prove that 
$$\tag{1} c m(D_j)\le \mu(D_j)\le C m(D_j),\qquad \forall j.$$ 
To prove this we fix $j$ and we let $b$ and $B$ denote two concentric balls such that 
$$b\subset D_j\subset B.$$
We can arrange things so that the ratio $\frac{m(B)}{m(b)}$ is a constant, not depending on $j$. Since 
$$\mu(b)\le \mu(D_j)\le \mu(B), $$
we infer 
$$C_1 m(b)\le \mu(D_j)\le C_2m(B).$$
It is also true that
$$m(b)\le m(D_j)\le m(B), $$
so we have for the ratio $\mu(D_j)/m(D_j)$ 
$$C_1 \frac{m(b)}{m(B)}\le \frac{\mu(D_j)}{m(D_j)}\le C_2\frac{m(B)}{m(b)}.$$
This is (1) with $c=C_1 \frac{m(b)}{m(B)}$ and $C=C_2\frac{m(B)}{m(b)}.$
