Define $f(x) = \dfrac{1}{x (\log x )^2}$ in $x \in (0,1/2)$.
Maximal function of $f$ is $$(Mf)(x) := \sup_{r>0} \frac{1}{m(B_r)} \int_{B(x,r)} |f(y)|\,dy,$$ where $B(x,r)$ denotes a ball of radius $r$ and center at $x$, and $m$ is a Lebesgue measure.
I tried to show that $Mf \notin L^1(\mathbb{R)}$. So I took a lower bound of $Mf$ like this:
Since $f(x) \geq e^2/4$ where $x \in (0,1/2)$,
\begin{align*} (Mf)(x) &= \sup_{r>0} \frac{1}{m(B_r)} \int_{B(x,r)} |f(y)|dy \\ & \geq \sup_{r>0} \frac{1}{m(B_r)} \int_{B(x,r)} e^2/4 dy \\ &= e^2/4 \end{align*} Therefore, $$\int_{\mathbb{R}} (Mf)(y) dy \geq \int_\mathbb{R} e^2/4 dy = \infty.$$
But I feel some weirdness about my solution now and cannot find where I exactly wrong. Does my solution make sense? If not, where is my mistake?
There is a general result on the maximal function which goes as follows: If $f\in L^1(\Bbb R)$ with $c:=\int_{B(0,R)}|f(y)|dy>0$ for some $R>0$, then $Mf\notin L^1(\Bbb R)$. This is, because for $|x|>R$ we have $$ Mf(x)\geq\frac{1}{\mu(B(x,R+|x|))}\int_{B(x,R+|x|)}|f(y)|dy\geq\frac{c}{\mu(B(x,R+|x|))}=\frac{c}{2(R+|x|)}, $$ and the expression on the right hand side is not integrable on $\Bbb R$.