Convergence and order

Question

Let $f \in L^1(\mathbb{R}^n)$, and $B_m = \left\{x \in \mathbb{R}^n \ | \ |x|< \frac{1}{m}\right\}$.

Then $\int_{B_m} |f| = O\left(\frac{1}{m^\delta}\right)$ as $m \to \infty$ for some $\delta >0$ ?

Answer 1

It is true that for an integrable function $f$, the sequence $\left(\int_{B_m} |f|\right)_{m\geqslant 1} $ converges to zero. This can be seen by approximating $f$ by a simple function. However, the decay can be arbitrarily slow. Indeed, let $n=1$ and let $\left(\delta_i\right)_{i\geqslant 1}$ be a decreasing sequence which converges to $0$. Let $$ f(x)=\sum_{i=1}^{+\infty} \frac{\delta_i-\delta_{i+1}}{i(i+1)}\mathbf 1_{\left[\frac 1{i+1},\frac 1i \right) }(x). $$ Then $f$ is integrable and $$\int_{B_m} |f|=\int \sum_{i=m}^{+\infty} \frac{\delta_i-\delta_{i+1}}{i(i+1)}\mathbf 1_{\left[\frac 1{i+1},\frac 1i \right) }(x)\mathrm d\lambda(x)=\delta_m.$$