If $\int_{\Omega\cap B_R}|f|\,dx\leq K R^n$ for some constant $K$, does $f\in L^\infty(\Omega)$?

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!

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

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 $$x \in \mathbf R^n$$.