Solutions of $u(x)=\int_{\mathbb R^n} |x-y|^p u(y)^{-q} dy$ are bounded away from zero In one of research papers I am interested in, see this link or this if you cannot access, there is a lemma, Lemma 5.1, saying that if $n\geq 1$, $p,q>0$ and $u$ a non-negative Lebesgue measureable function which solves the following integral equation
$$u(x)=\int_{\mathbb R^n} |x-y|^p u(y)^{-q} dy$$
in the whole space $\mathbb R^n$, then there is a constant $C>0$ such that
$$u(x) \geqslant \frac {1+|x|^p}C.$$
The proof goes as follows: By means of the equation, $u$ must be positive everywhere and
$$meas(\{y \in \mathbb R^n: u(y)<+\infty\})>0.$$
Therefore, there exists some $R>1$ and some measurable set $E \subset \mathbb R^n$ such that
$$E \subset \{y \in \mathbb R^n: u(y)<R\} \cap B(0,R), \quad meas(E) \geqslant \frac 1R.$$
Consequently,
$$u(x) \geqslant \int_E |x-y|^p u(y)^{-q} dy \geqslant R^{-q}\int_E |x-y|^p dy$$
for all $x \in \mathbb R^n$. From this the assertion follows.
To me, the last step is not clear, it seems that the author has already used the fact $ meas(E) \geqslant \frac 1R$ somewhere to get the lower bound.
Please advise. Thank you.
 A: Observe that $\left|\left\{u<+\infty\right\}\right|\neq 0$. Otherwise, $u^{-q}=0$ almost everywhere and therefore
$$u(x)=\int_{\mathbb{R}^{n}}\left|x-y\right|^{p}u(y)^{-q}=0$$
on $\mathbb{R}^{n}$. Also, $u$ cannot vanish on $\mathbb{R}^{n}$. Suppose $u$ vanishes at $x_{0}$, and let $B(x_{0},\delta)$ be an open ball of radius $\delta\ll 1$ centered at $x_{0}$. Then
$$u(x_{0})\geq\int_{\left|y-x_{0}\right|\geq\delta}\left|x_{0}-y\right|^{p}u(y)^{-q}\geq\delta^{p}\int_{\left|y-x_{0}\right|\geq\delta}u(y)^{-q}>0,$$
for $\delta$ sufficiently small, since $u^{-1}>0$ on a set of positive measure as previously noted. From $\sigma$-finiteness and continuity of measure, we see that there exists an $R>1$ and a measurable set $E$ such that
$$E\subset\left\{u<R\right\}\cap B(0,R), \quad \left|E\right|\geq R^{-1}$$
Using the definition of $E$, we obtain that
$$u(x)\geq\int_{E}\left|x-y\right|^{p}u(y)^{-q}dy\geq R^{-q}\int_{E}\left|x-y\right|^{p}dy,\qquad\forall x\in\mathbb{R}^{n}$$
By dominated convergence, the function $\phi: x\mapsto\int_{E}\left|x-y\right|^{p}dy$ is continuous on $\mathbb{R}^{n}$. For $\left|x\right|>2R$, we have by the reverse triangle inequality that
$$\phi(x)\geq\int_{E}(\left|x\right|-\left|y\right|)^{p}dy\geq\int_{E}(2R-R)^{p}dy=R^{p}\left|E\right|\geq R^{p-1}$$
The closed ball $\overline{B}(0,2R)$ is compact, so by Weierstrass' extreme value theorem, $\phi$ attains its minimum at some point $x_{0}\in\overline{B}(0,2R)$. 
$$\phi(x_{0})=\sum_{k\in\mathbb{Z}}\int_{E\cap\left\{2^{k}\leq\left|x_{0}-y\right|<2^{k+1}\right\}}\left|x_{0}-y\right|^{p}dy\geq\sum_{k\in\mathbb{Z}}2^{kp}\left|E\cap\left\{2^{k}\leq\left|x_{0}-y\right|<2^{k+1}\right\}\right|>0,$$
since $\left|E\right|>0$.
For $x\neq 0$, set $E_{k}:=E\cap\left\{2^{k}\left|x\right|\leq\left|y\right|<2^{k+1}\left|x\right|\right\}$. Observe that
$$\begin{array}{lcl}\displaystyle\int_{E}\left|x-y\right|^{p}dy\geq\int_{E}\left|\left|x\right|-\left|y\right|\right|^{p}dy&=&\displaystyle\sum_{k\in\mathbb{Z}}\int_{E_{k}}\left|\left|x\right|-\left|y\right|\right|^{p}dy\\[2 em]\displaystyle&\geq&\displaystyle\sum_{k=0}^{\infty}(2^{k}\left|x\right|-\left|x\right|)^{p}\left|E_{k}\right|+\sum_{k=-1}^{-\infty}(\left|x\right|-2^{k}\left|x\right|)^{p}\left|E_{k}\right|\\[2 em]&=&\displaystyle C'\left|x\right|^{p}\end{array}$$
where $0<C'<\infty$ is a constant which does not depend on $x$. Set $C:=2^{-1}\min\left\{\phi(x_{0}),C'\right\}$. Putting all our results together, we conclude that
$$u(x)\geq R^{-q}\int_{E}\left|x-y\right|^{p}dy\geq CR^{-q}\left(1+\left|x\right|^{p}\right),\qquad\forall x\in\mathbb{R}^{n}$$
