Convergence rate for $\int_{\{f\ge n\}}|f(x)|\,dx$ for $f\in L^1(\Omega)$

Question

Let $\Omega\subset\mathbb R^d$ be open and bounded and $f\in L^1(\Omega)$. Consider the following sequence indedxed by $n\in\mathbb N$: $$ \int_{\{f\ge n\}}|f(x)|\,dx .$$ This sequence converges to zero by the Beppo Levi-theorem. My question: Can one find an explicit convergence rate which is independent of $f$? More specifically, I'm looking for something along the lines of $$\int_{\{f\ge n\}}|f(x)|\,dx\leq C_f n^{-\alpha},$$ with some $\alpha>0$ and a constant $C_f$ which is allowed to depend on $f$.

Answer 1

(this is only a partial answer)

consider $\Omega=(0,1)$, $f=x^\beta, 0>\beta>-1$.

We have $f \geq n \Leftrightarrow x \leq n^{\beta^{-1}}$.

Then $$ \int_{f\geq n} |f| = \int_0^{n^{\beta^{-1}}} x^\beta \mathrm dx = \frac{{(n^{\beta^{-1}})}^{\beta+1}}{\beta+1} = \frac{n^{1+\beta^{-1}}}{\beta+1} $$

Because $\beta$ can be arbitrarily close to $-1$, the value $\alpha:=-1-\beta^{-1}$ can be arbitrarily small.

So the convergence rate is worse than $n^{-\alpha}$ for every $\alpha>0$, when $\alpha$ has to be independent of $f$.