Let me restate Vitali covering lemma.

Let $\{B_i\}_{i\in F}$ be a finite collection of balls in the $\mathbb{R}^n$. Then there is $S\subset F$ such that the balls $\{B_i\}_{i\in S}$ are disjointed and

$$\mathop{\rm vol}\left(\bigcup_{i\in S}B_i\right)\ge \tfrac1{3^n} \mathop{\rm vol} \left(\bigcup_{i\in F}B_i\right),$$

Question. Can one make the constant better? In particular, is it OK to change $\tfrac1{3^n}$ to $\tfrac1{2^n}$ in the formulation?

It is likely that the answer is well known, but googling did not help.

Clearly the greedy algorithm which is used in the standard proof of Vitali covering lemma can not give anything better than $\tfrac1{3^n}$.

One can not make it better than $\tfrac{1}{2^n}$. (Consider a large collection of balls of the same radius such that each contains a fixed point.)

I see that for $n=1$ the optimal constant is $\tfrac1{2}$, but for higher dimensions I do not see a proof.

Let me restate Vitali covering lemma.

Let $\{B_i\}_{i\in F}$ be a finite collection of balls in the $\mathbb{R}^n$. Then there is $S\subset F$ such that the balls $\{B_i\}_{i\in S}$ are disjoint and

$$\mathop{\rm vol}\left(\bigcup_{i\in S}B_i\right)\ge \tfrac1{3^n} \mathop{\rm vol} \left(\bigcup_{i\in F}B_i\right),$$

Question. Can one make the constant better? In particular, is it OK to change $\tfrac1{3^n}$ to $\tfrac1{2^n}$ in the formulation?

It is likely that the answer is well known, but googling did not help.

Clearly the greedy algorithm which is used in the standard proof of Vitali covering lemma cannot give anything better than $\tfrac1{3^n}$.

One cannot make it better than $\tfrac{1}{2^n}$. (Consider a large collection of balls of the same radius such that each contains a fixed point.)

I see that for $n=1$ the optimal constant is $\tfrac1{2}$, but for higher dimensions I do not see a proof.