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Let $\Omega\subset\mathbb{R}^n$ be a smooth bounded domain. Assume that for a measurable function $f$, there exists a constant $K>0$ such that $$ \int_{\Omega\cap B_R}|f|\,dx\leq K R^n, $$ where $B_R$ is a ball. My question is that how to prove $f\in L^\infty(\Omega)$. If $f$ is continuous, then this proposition is trivial. For general $f$, I don't know how to prove it. Actually, I know there exists a set sequence $\{\Omega_n\}$ such that $|f(x)|>n$ in $\Omega_n$. But since $f$ is not continuous and $\Omega_n$ is just measurable, we can't find a ball $B_n\subset\Omega_n$ (actually I'm not sure about this, maybe we can let $\Omega_n$ be open but I don't know how to do it), so we can't get a contradiction.

I would appreciate it if anyone could give me some help. Thanks!

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    $\begingroup$ If your inequality holds for all $R$, then appying Lebesgue differentiation theorem to $|f|\chi_{\Omega}$ may help. $\endgroup$
    – Feng
    Jul 18 at 13:04
  • $\begingroup$ oh thank you!!! This really helps me a lot!!! $\endgroup$
    – Kimura Leo
    Jul 18 at 13:43

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Extend $f$ by $0$ off the domain $\Omega$ (neither smoothness nor boundedness is required). Then you have $$\frac 1{|B_R|} \int_{B_R} |f| \, dx \le \frac K{\omega_n}$$ where $\omega_n$ is the $n$-volume of the unit ball. Since this holds for every ball, you have that $$\limsup_{R \to 0^+} \frac 1{|B_R(x)|} \int_{B_R(x)} |f| \, dx \le \frac K{\omega_n}$$ for every $x \in \mathbf R^n$. According to the Lebesgue Differentiation Theorem you must have $$|f(x)| \le \frac K{\omega_n}$$ for almost every $x \in \mathbf R^n$.

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  • $\begingroup$ Oh my, this theorem is very powerful! Thanks for your help!!! $\endgroup$
    – Kimura Leo
    Jul 18 at 13:45

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