We prove that for any weight $\phi$ defined on $[0,1]^n$ that satisfies areverse Holder inequality with exponent p > 1 and constant $c\ge1$ upon alldyadic subcubes of $[0,1]^n$, it's non increasing rearrangement satisfies areverse Holder inequality with the same exponent and constant not more than$2^nc-2^n + 1$, upon all subintervals of $[0; 1]$ of the form $[0; t]$. Thisgives as a consequence, according to the results in [8], an interval $[p;p_0(p; c)) = I{p,c}$, such that for any $q \in I{p,c}$, we have that $\phi$ isin $L^q$.
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机译：我们证明，对于在$ [0,1] ^ n $上定义的任何权重$ \ phi $，满足满足以下条件的Holder不等式：指数p> 1且对于$ [0,1] ^ n的子对子立方体为常数c $，它的非递增重排满足在$ [0; 0]的所有子间隔上具有相同指数且常数不超过$ 2 ^ nc-2 ^ n + 1 $的反向Holder不等式。 1] $，格式为$ [0; t] $。结果，根据[8]中的结果，间隔$ [p; p_0（p; c））= I {p，c} $，使得对于$ {\ in I {p，c} $，我们有$ \ phi $ is $ L ^ q $。
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