Possible mistake in Safonov's lemma 
$\newcommand{\measure}[1]{\lvert#1\rvert}$Let $B_{R_0}$ a ball in $\mathbb{R}^d$, $R_{0}>0, 0<\xi<1$ and $\Gamma \subset B_{R_0}$ be a  measurable set such that $\measure\Gamma >0$. Denote
  $$K:=\{B-ball:B\subset B_{R_0} \ \  \mbox{and} \ \ \measure{B\cap \Gamma} \ge \xi \measure B\}
\ \ \mbox{and } \ \ 
\Gamma_0:= \textstyle\bigcup_{B \in K} B. $$
  If $\measure\Gamma < \xi \measure{B_{R_0}}$, then $\measure{\Gamma_0} \ge (1+\nu)\measure\Gamma$
  where $ \nu=\frac{1-\xi}{3^d}>0$.

This is the content of Lemma 1.1 of "Harnack's inequality for elliptic equations and the Hölder property of their solutions"
by M. V. Safonov.
However I was thinking the following. In this situation,
\begin{equation}
\measure{B_{R_0}} \ge \measure{\Gamma_0}  \ge (1+\nu)\measure\Gamma= \left  \{ (1+\nu)\dfrac{\measure\Gamma}{\measure{B_{R_0}}} \right  \} \measure{B_{R_0}}.
\end{equation}
Thus in order we do not have a contradiction we need to have
\begin{equation}
(1+\nu)\dfrac{\measure\Gamma}{\measure{B_{R_0}}} \le 1,
\end{equation}
that is
\begin{equation}
\frac{1-\xi}{3^d}= \nu \le \dfrac{\measure{B_{R_0}}}{\measure\Gamma}-1.
\end{equation}
Equivalently,
$$\xi \ge 1- 3^d \left ( \dfrac{\measure{B_{R_0}}}{\measure\Gamma}-1 \right ).  $$
We can see that if $\measure\Gamma$ is sufficiently close to $\measure{B_{R_0}}$ the right-hand-side above is close to "1". My question then is the following. Am I wrong or is the Lemma 1.1 is false in the sense that this is not true for arbitrary $\xi$? And the statement could be

Let $B_{R_0}$ a ball in $\mathbb{R}^d$, $R_{0}>0$ and $\Gamma \subset B_{R_0}$ be a measurable set such that $\measure\Gamma >0$. Denote
  $$K:=\{B-ball:B\subset B_{R_0} \ \  \mbox{and} \ \ \measure{B\cap \Gamma} \ge \xi \measure B\}
\ \ \mbox{and } \ \ 
\Gamma_0:= \textstyle\bigcup_{B \in K} B. $$
  If $\measure\Gamma < \xi \measure{B_{R_0}} $ with $0<\xi<1$ such that 
  $$\xi \ge 1- 3^d \left ( \dfrac{\measure{B_{R_0}}}{\measure\Gamma}-1 \right ),  $$
  then $\measure{\Gamma_0} \ge \left(1+\frac{1-\xi}{3^d}\right)\measure\Gamma?$


I'm afraid you're wrong here; the lemma is fine as it is. The hypothesis of the lemma, $\lvert \Gamma \rvert < \xi \lvert B_{R_0} \rvert$, or more suggestively $\xi > \lvert \Gamma \rvert / \lvert B_{R_0} \rvert$, is a stronger condition than the inequality that you derived. To check, let $x = \lvert \Gamma \rvert / \lvert B_{R_0} \rvert$. Then $x<1$ implies $(x-1)(x-3^d) > 0$, which is equivalent to $$x^2 > x - 3^d(1-x).$$ Since $x>0$, dividing the above inequality by $x$ yields $x > 1 - 3^d(x^{-1} - 1)$, as desired.
