Application of weak $L^p$ estimate besides for proving boundedness of some linear operator For all $1\leq p< \infty$, weak-$L^p(\mathbb{R}^d)$ space is defined as a set  of all functions $f$ such that 
$$\gamma^p|\{x\in \mathbb{R}^d: |f(x)|>\gamma\}|<\infty$$
for every $\gamma>0$.
By Chebyshev inequality, if $f\in L^p(\mathbb{R}^d)$, then $f\in$ weak-$L^p(\mathbb{R}^d)$.
Let $T$ be a linear operator on  weak-$L^p(\mathbb{R}^d)$ and $1\leq p <\infty$.
If there exist a constant $C_1>0$ such that
$$\gamma^p|\{x\in \mathbb{R}^d: |Tf(x)|>\gamma\}|\leq C_1 \|f\|_p^p.$$
for every $\gamma>0$, 
then by using Marcinkiewicz Interpolation Theorem, we have the boundedness of $T$ on $L^p(\mathbb{R}^d)$ for $1<p<\infty$, that is $\|Tf\|_p\leq C_2\|f\|_p$.
One of the examples of the operator $T$ is Hardy-Littlewood maximal operator $M$.  
My question is that what is the application of weak type estimates of linear operator besides proving the boundedness of the operator? 
 A: Some important operators are not bounded on $L^1$, like the Hilbert transform or the Hardy-Littlewood maximal operator. But they satisfy a weak $(1,1)$ bound, which is good enough to run some density arguments. By density, I mean that a given $f\in L^1$ can be written as $g+h$ where $g$ is nice (smooth) and $h$ has small $L^1$ norm. Then we use  the weak $(1,1)$ bound to control the influence of $h$. We cannot say that the contribution of $h$ is small in $L^1$ norm, but we know that it is small outside of a set of small measure.
For example: one proves the Lebesgue differentiation theorem for $f\in L^1$  by writing $f=g+h$ as above, observing that $g$ presents no difficulties, and that $Mh$ is small on most of the space. Since $Mh$ controls the effect of $h$ on the Lebesgue derivative, this leads to the conclusion that a.e. point is a Lebesgue point of $f$.  
Or take the Hilbert transform $H$.  Due to the singular kernel, it is not obvious how to define $Hf$ for a general $f\in L^1$. But we can define it for smooth $g$, and then use the weak $(1,1)$ bound to conclude that $Hf$ is a well-defined function (a.e.), and in fact belongs to the weak $L^1$ space.
