I am not sure if this is a standard theorem, since I can't find a name for it: Suppose that $f\in L^1(\mathbb R, m)$ where $m$ is the standard lebesgue measure. Then for every $\epsilon >0$, there is an $R>0$ such that$$\int_{|x|>R} |f| <\epsilon$$
I tried proving it in the following manner:
Suppose there is an $\epsilon >0$ such that for all $R>0$,$$\int_{|x|>R} |f| \geq \epsilon$$
Note that$$|f(x)|\mathbb{1}_{ \{|x|>\} R} (x) \to 0 \text{ as } R\to \infty$$almost anywhere since $\{|x|=\infty\}$ is a null set (it's a singleton). By the Dominated Convergence Theorem, we get that
$$\lim_{R\to \infty} \int_{\{|x|>R\}} |f(x)| dx = 0 \geq \epsilon$$yielding a contradiction.
I would like to know if this result has a standard name and whether my proof needs any revision?