If $(f_n)$ measurable on $(X,\mathcal{M},\mu)$, $f_n$ Cauchy with respect to weak quasi $L^p$-norm:
$[f_n]_p=\sup_{\alpha>0}\alpha \lambda_{f_n}(\alpha)^{\frac{1}{p}} $
where $\lambda_{f}(\alpha)=\mu(\{x:\vert f(x)\vert > \alpha \}) $, is it necessarily true that $(f_n)$ is Cauchy with respect to $\vert \vert \cdot \vert \vert_p$?
First of all it is possible that a sequence can be convergent in the sense of the weak-$\mathbb L^p$-norm (defined by $\left\lVert f\right\rVert_{p,\infty}=\sup_{A:\mu(A)>0}\mu(A)^{-1+1/p}\int_A \left\lvert f\right\rvert d\mu$ in the case of finite measure) such that none of elements of the sequence are in $\mathbb L^p$.
And even if it is the case, there are counter-example. We work on a probability space and consider a random variable $X\geqslant 0$ such that $2^{kp}\mathbb P\left(X\geqslant 2^k\right)\sim c/k$. Let $X_n=X\mathbf 1\{X\leqslant n\}$. Then we can check that $X_n\to X$ in the sense of weak-$\mathbb L^p$-norm hence $(X_n)$ is Cauchy but it cannot be Cauchy in the complete space $\mathbb L^p$, otherwise it would be convergent to $X$, which is not in this space.