Show that for every $\epsilon >0$ exist a compact set $K\subset \mathbb R^n$ such that $\int_{\mathbb R^n \setminus K}|f|d\lambda_n<\epsilon$ 
Let integrable function $f: \mathbb R^n \to \mathbb R^n$ . Show that for every $\epsilon >0$ exist a compact set $K\subset \mathbb R^n$ such that $\int_{\mathbb R^n \setminus K}|f|d\lambda_n<\epsilon$

Unfortunately, I could not think of anything special that would lead me to the proof ... Can I have some tips?
 A: Let $B_r$ be the closed ball of radius $r$ centered at $0$. Then 
$$\lim_{r\to\infty}\int_{B_r}|f|d\lambda_n = \int_{\mathbb R^n}|f|d\lambda_n$$
So there exists $t > 0$ such that
$$\left|\int_{\mathbb R^n}|f|d\lambda_n -\int_{B_t}|f|d\lambda_n \right| \le \epsilon$$
but the expression on the left equals $\int_{\mathbb R^n\backslash B_t}|f|d\lambda_n$.
A: Let $K_j$ be the closed ball of radius $j$, centered at the origin, and le $g_j$ be the functions defined by
$$
g_j(x)=|f(x)| \textrm{ if } x\in K_j,\quad g_j(x)=0 \textrm{ if } x\in \mathbb R^n\setminus  K_.
$$
Moreover,  let $I_j=\int_{\mathbb R^n}g_j d\lambda_n$.
Then,  $g_n$ monotonically converges to $|f|$ and by the monotone convergence theorem,
$$
I_j\to I:=\int_{\mathbb R^n}|f|d\lambda_n<\infty,
$$
with $I_j\le I_{j+1}$. 
Since  the increasing sequence $\{I_j\}$ is convergent, for any $\varepsilon>0$ there is an index $\ell$ such that
$$
0\le I-I_j<\varepsilon\quad \forall j\ge \ell.
$$
Then you conclude that
$$
\int_{\mathbb R^n\setminus I_\ell}|f|\le \varepsilon.
$$
