Let $f$ be a measurable function in the extended real topology $[-\infty, \infty]$, $f:\mathbb{R}^n\to [-\infty, \infty]$. Then $f$ is said to belong to the weak $L^p(\mathbb{R}^n)$ for $1 \leq p < \infty$ if there exists a global constant $M \geq 0$ such that forall $c > 0$ we have $c^p\mu\left(\{x \in \mathbb{R}^n: |f(x)| > c\}\right)\leq M$.
I am trying to show that every $f \in L^p(\mathbb{R}^n)$ also belongs to the weak $L^p(\mathbb{R}^n)$. One approach is to take $M = ||f||_{L^p(\mathbb{R}^n)}^p < \infty$ and to observe that
$$||f||_{L^p(\mathbb{R}^n)}^p = \int_{\mathbb{R}^n}|f|^pdx \geq \int_{\{x\in \mathbb{R}^n: |f(x)| > c\}}|f|^pdx > c^p\mu\left(\{x \in \mathbb{R}^n: |f(x)| > c\}\right)$$
for all $c > 0$. But as of writing, it is unclear to me how you can argue about the equality, when we now only have a strict inequality.
Edit: It is my fault that I did not communicate my problem clearly. My problem is that I am not sure whether the approach I mentioned shows the possibility of an equality. IMO we only have a strict inequality.