Let $B(x,R)$ denotes the ball in centered at $x\in \mathbb{R}^n$ with radius $R$. The centered Hardy-Littlewood maximal operator $M$ is defined by \begin{equation} Mf(x)=\sup_{B(x,R)} \frac{1}{|B(x,R)|} \int_{B(x,R)} |f(y)| dy \end{equation} for every locally integrable function $f \in L^1_{\text{loc}}(\mathbb{R}^n)$. We know that $M$ is of weak type-$(1,1)$, that is, there exist a constant $C>0$ such that \begin{equation} |\{ x\in \mathbb{R}^n: Mf(x)>\lambda \}| \le \frac{C\|f\|_{L^1}}{\lambda}, \end{equation} for every $\lambda>0$ and $f\in L^1(\mathbb{R}^n)$.
My question is that could we improve the right hand side bound in the sense as follows:
There exist a constant $C>0$ such that \begin{equation} |\{ x\in \mathbb{R}^n: Mf(x)>2\lambda \}| \le \frac{C\|f \chi_{\{|f|>\lambda\}}\|_{L^1}}{\lambda}, \end{equation} for every $\lambda>0$ and $f\in L^1(\mathbb{R}^n)$.
Yes, this result is true. And actually you can explicitly write, for $f$ non-negative, $$\mu\{Mf>\alpha\}\leq \frac{3^d}{(1-\delta)\alpha}\int_{\{f>\delta\alpha\}}f\,dx $$ where $d$ denotes the dimension.
For general $f$, write $f=f^+-f^-$
Let $E_\alpha:=\{Mf > \alpha \}$. By definition of maximal function we can choose a ball $B_x$, centered at $x$ for each $x\in E_\alpha$, such that \begin{equation} \frac{1}{\mu(B_x)}\int_{B_x}|{f}|dx>\alpha, \end{equation} and $\{B_x\}_{x\in E_\alpha}$ covers $E_\alpha$.
Then by Vitali covering theorem, we can choose a sequence of disjoint balls $\{B_k\}$ from $\{B_x\}_{x\in E_\alpha}$ such that $E_\alpha\subset \bigcup_k B_k$ and
$$\mu(E_\alpha)\leq 3^d \sum_k \mu(B_k).$$
Then by we have \begin{equation} \mu(E_\alpha)\leq 3^d\sum_k \mu(B_k) \leq \sum_k\frac{3^d}{\alpha}\int_{B_k}|{f}|dx\leq \frac{3^d}{\alpha}\int_{\bigcup_kB_k}|{f}|dx \end{equation} Now since $f\geq 0$, for each given $\alpha$, we can choose $0<\delta<1$ such that
$$\{Mf(x)>\alpha\}\subset \bigcup_k B_k\subset \{f(x)>\delta\alpha\}.$$ Thus, we have \begin{equation} \mu(E_\alpha)\leq \frac{3^d}{\alpha}\int_{\bigcup_kB_k}|{f}|dx \leq \frac{3^d}{\alpha}\int_{\{{f>\delta\alpha}\}}|{f}|dx\leq \frac{3^d}{\alpha(1-\delta)}\int_{\{f>\delta\alpha\}}|{f}|dx. \end{equation}