Let $f \in L_{loc}^1(\mathbb{R}^n)$ (i.e. $f$ restricted to any compact set is in $L^1$) and $\mu$ be the Lebesgue measure. I‘m trying to show that $f^*(x):=\sup_{r>0} \int_{B_r(x)}|f|d \mu/ \mu(B_r(x))$ is not in $L^1(\mathbb{R}^n)$ unless $f=0$ a.e. .
By contradiction I assumed that $\mu(A) \neq 0$ for $A:=\{x \in \mathbb{R}^n | f(x) \neq 0 \}$. Now if $A$ were bounded I can find an $R>0$ s.t. $A \subseteq B_R(0)$. Then for all $x \in \mathbb{R}^n$ s.t. $|x| > R$ we have $f^*(x) \geq c|x|^{-n}, \ c >0$ because:
$$ \sup_{r>0} \int_{B_r(x)}|f|d \mu/ \mu(B_r(x)) \geq \int_{A}|f|d \mu/ \mu(B_{|x|+R}(x)) \geq \int_{A}|f|d \mu/ \mu(B_{2|x|}(x)) \geq c/|x|^n $$
And thus $f^* \notin L^1(\mathbb{R}^n) $. Am I correct thus far?
But I dont know how to proceed in the general case, i.e. if $A$ isn’t bounded, any help would be appreciated:)
Since $f\in L^1(\mathbb{R}^n)$, most of the integral is obtained in some sufficiently large ball. I.e. given $\varepsilon>0$, there exists $R>0$ such that $\int_{\mathbb{R}^n\backslash B_R(0)}|f|d\mu<\varepsilon$. Then you can make a similar argument to what you have already, choosing $0<\varepsilon<\|f\|_{L^1(\mathbb{R}^n)}$ arbitrarily one has for some $R>0$ sufficiently large: $$ \sup_{r>0}\frac{1}{\mu(B_r(x))}\int_{\mathbb{R}^n}|f|d\mu\geq\frac{1}{\mu(B_R(x))}\int_{B_R(x)}|f|d\mu\geq \frac{\|f\|_{L^1(\mathbb{R}^n)}-\varepsilon}{\mu(B_{R+|x|}(0))}\geq\frac{\|f\|_{L^1(\mathbb{R}^n)}-\varepsilon}{C(R+|x|)^n}\sim O(|x|^{-n})$$ Since this can be obtained for any such $\varepsilon$, we see this quantity decays at most as fast as $|x|^{-n}$, and so cannot be in $L^1(\mathbb{R}^n)$.