Let $f\in L^p$. Can we say $\|f\|_{L^{p}} \leq \epsilon$ on $|x|\geq R$ for large $R$? Let $f\in L^{p}(\mathbb R), (1\leq p &lt;\infty)$ and $\epsilon&gt;0.$ 


  My Question: Can we expect  to find $R&gt;1$ (may be large) so  that
   $\|f\|_{L^{p}(B_R)} \leq \epsilon$  on $B_{R}=\{x:|x|\geq R\}$?
The function $ g_n = \chi_{[-n,n]} |f|^p$ satisfies 

$$0\le g_1 \le g_2\le \cdots \le |f|^p$$

and $g_n \to |f|^p$ pointwisely on $\mathbb R$. Thus the monotone convergence theorem implies that 

$$\left( \int_{-n}^n |f|^p\right)^{\frac{1}{p}} \to \|f\|_p.$$

If $\int_\mathbb{R} |f|^p$ is finite, then for all $\epsilon &gt;0$ there is $n$ so that 

$$ \|f\|^p_p -\epsilon^p \le \int_{-n}^n |f|^p \le \|f\|^p_p,$$

which implies 

$$\left(\int_{|x|\ge n} |f|^p\right)^{\frac 1p} &lt;\epsilon.$$
Let $f$ consist of small non-overlapping rectangular bumps of height 1
centered on 1,2,3,$\cdots $,n,$\cdots $ with width $w_{n}=1/n^{k}$ with
sufficiently large $k$. Then $f\in L^{p}$ but will never become smaller than
$\varepsilon $ as $x$ increases.