Non-existence of bounded linear functional on weak-type space

I'm studying the functional analysis book of Stein and Shakarchi. I have a trouble in solving some exercises. The question is about showing non-existence of bounded linear functional on weak-type space.

Specifically, we define the weak-type space as a set of function for which $m(\{ x : |f(x)| > \alpha \}) \leq \frac{A}{\alpha}$ for some $A$ and all $\alpha > 0$. Also, we define the quantity $\mathcal{N}(f)$ as the infimum of $A$ that satisfying the inequality above(although this quantity is not a norm).

The exercise I have a trouble with is showing that this space(weak-type space) has no non-trivial bounded linear functional. To show this, I have assumed the existence of bounded linear functional $l$ and want to show actually $l(f) = 0$ for all function $f$ of weak-type space. The followings are what I have tried.

By assumption, we have some $M > 0$ with $|l(f)| \leq M \mathcal{N}(f)$ for all $f$ so $\mathcal{N}(f) \geq \frac{1}{M}|l(f)|$. I made some observation : $\mathcal{N}(f) > \beta$ is equivalent with existence of some $\alpha > 0$ where $m(\{x : |f(x)| > \alpha\}) > \frac{\beta}{\alpha}$. With the observation, for all $f$, there exists $\alpha > 0$ such that $m(\{|f(x)| > \alpha \}) > \frac{1}{M\alpha}|l(f)|$ or $\mathcal{N}(f) = \frac{|l(f)|}{M}$. Now I have no progress in this problem. Could you give me some hint or direction to solving this problem please?