Consider the maximal operator defined for a function $f\in L^1_{loc}$  : $Mf : x\mapsto \sup_{r>0} \frac{1}{|B(x,r)|} \int_{B(x,r)} f$.
It is well know that $M : L^1 \to L^{(1,\infty)}$ and $M : L^p \to L^p$ if $p>1$.
But now, if for some function $f$, we have $Mf\in L^1$, then, it is also known that $f\in L^1\log L$.
My question is about what if we impose slightly more on the maximal function : Assume $Mf\in L^1 \log^t L$ for some $t\geq 0$, can we prove that $f$ is better than $L^1 \log L$ ? For instance $L^1 \log^{1+t} L$ ?

Consider the maximal operator defined for a function $f\in L^1_{loc}$: $$ Mf : x\mapsto \sup_{r>0} \frac{1}{|B(x,r)|} \int\limits_{B(x,r)} f. $$ It is well know that $M : L^1 \to L^{(1,\infty)}$ and $M : L^p \to L^p$ if $p>1$.
But now, if for some function $f$, we have $Mf\in L^1$, then, it is also known that $f\in L^1\log L$.
My question is about what if we impose slightly more on the maximal function : Assume $Mf\in L^1 \log^t L$ for some $t\geq 0$, can we prove that $f$ is better than $L^1 \log L$ ? For instance $L^1 \log^{1+t} L$ ?