I apologize if my notation is not 100% correct. I'm not even sure if this fact requires proof or is considered obvious.
If $f$ is integrable, i.e. $\int_{\mathbb{R}}|f| < \infty$, does it follow that $\mu \{x \in \mathbb{R}:|f|>t \} \to 0$ as $t \to \infty$. Here's my proof:
For $t$, define sequence $t_0 < t_1 \ldots <t_n, n \to \infty$ and a sequence of disjoint intervals $I^{t_n}_j = [x_j, x_{j+1}]$ equipped with measure $\mu$ such that $f>t_n$ on each of these intervals. by countable additivity and monotonicity of $t_n$ (again, I'm not 100% sure of this step - is this enough or the inequality requires a deeper proof?), $$ \mu (\cup_{j=1}^{\infty}I_j) = \sum_{j=1}^{\infty}\mu(I^{t_{n}}_j) \geq \sum_{j=1}^{\infty}\mu(I^{t_{n+1}}_j) $$ By integrability, if $\int |f|< \infty$, then $|f|<\infty$, and, since the sequence $t_n$ is infinite, $\exists \ n^{\ast}$ s.t. $t_{n^{\ast}} < f < t_{n^{\ast}+1}$ and $\mu(I^{t_{n}}_j)=0 \ \forall n>n^{\ast}+1$. Therefore, $$ \lim_{n \to\infty} \sum_{j=1}^{\infty}\mu(I^{t_{n+1}}_j) = \sum_{j=1}^{\infty}\lim_{n \to \infty}\mu(I^{t_{n+1}}_j)=\sum_{j=1}^{\infty}0=0 $$ Hence $\mu \{|f|>t \} \to 0$.