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I would be interested to know whether the following is true:

Let $\Omega$ be a bounded open set in $\mathbf{R}^n$. Let $g$ be a nonnegative function $g : \Omega \to \mathbf{R}$. If there is a constant $C > 0$ such that \begin{equation} \frac{1}{|A|^{1-1/p}} \int_A g \leq C \end{equation} for all measurable subset $A \subset \Omega$, then $g$ is in weak-$L^p(\Omega)$.

If the above inequality holds only for open balls in $\Omega$, is $g$ still in weak-$L^p(\Omega)$ ?

Edit: I changed the question to make it more relevant and less naive. Most comments below are out of date.

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    $\begingroup$ Is this a homework problem? Basically the question is whether weak $L_p$ and $L_p$ are the same. The answer is probably in your text book. $\endgroup$
    – Bill Johnson
    Commented Nov 6, 2011 at 23:57
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    $\begingroup$ ...and also $g(x)=\Vert x\Vert^{-n/p}$ on $B_1(0)\subseteq\mathbb{R}^n$. $\endgroup$
    – George Lowther
    Commented Nov 7, 2011 at 1:07
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    $\begingroup$ The answer is no for $p>1$. Consider a function distributed like $x^{-1/p}$ on $[0,t]$ for small $t$, and add together translates of this function so that the distance between the supports of the functions are big with respect to $t$. $\endgroup$
    – Bill Johnson
    Commented Nov 7, 2011 at 23:40
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    $\begingroup$ @unknown (google): No, it is not in weak $L^p$, but that is the point. It still satisfies the inequality for open balls. $\endgroup$
    – George Lowther
    Commented Nov 8, 2011 at 0:41
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    $\begingroup$ Moving targets can be difficult to hit... $\endgroup$
    – Yemon Choi
    Commented Nov 8, 2011 at 8:56

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