Let $f\in L^1(\mathbb R).$ Let $r$ be any real number and let $B_r:=\{x:|x|\le r\}.$
Then we know, $$\int_{\mathbb{R}\setminus B_r} |f(x)|dx\to0\,\text{ as }\,r \to \infty.$$
My question is, how to prove this by definition (i.e. without appealing to any theorem or fancy stuff)?
My method (involving dominated convergence theorem):
$\int_{\mathbb R}|f\chi_{B_r}|\le\int_{\mathbb R}|f|\lt \infty$ and $f\chi_{B_r}\to f$ pointwisely as $r\to\infty$, where $\chi_{B_r}$ is the characteristic function on $B_r.$ Then by the dominated convergence theorem, we are done.
If $f$ is an integrable simple function, say $\sum a_iI{E_i}$ ( where $E_i$'s are pairwise disjoint measurable sets) then the result follow from the fact that $\mu (E_i \setminus B_r) \to 0$. [Note that $\mu (E_i) <\infty$]. The general case follows from the fact that there exists a simple function $g$ with $\int |f-g| d\mu <\epsilon$. Note that $\int_{\mathbb R \setminus B_r} |f-g|d\mu <\epsilon$ for all $r$.