Show that $\exists x_i \in \mathbb{R}$ and $i \in \mathbb{N}$ such that $\lim_{i\to \infty} x_i = \infty$ and $\lim_{i \to \infty} x_i f(x_i) = 0$ 
Suppose that $f : \mathbb{R} \to \mathbb{R}$ is an integrable function. Show that there exists $x_i \in \mathbb{R}$ and $i \in \mathbb{N}$ such that $$\lim_{i\to \infty} x_i = \infty \text{ and } \lim_{i \to \infty} x_i f(x_i) = 0.$$

Supposing the contrary one has that $\lim_{i\to \infty} x_i = 0$ and $\lim_{i\to \infty} x_if(x_i) = \infty$. From integrability of $f$ one has also that $\int_{\mathbb{R}} |f|< \infty$. The assupmtion seems to imply that $x_if(x_i)$ diverges faster than $x_i$ converges. Somehow this feels like the assumption that the integral is finite would become a problem if $\lim_{i\to \infty} x_if(x_i) = \infty$, but I don't really see how? Is there a direct way to approach this or is the contradiction the way to go?

Let $A=\{x\in(1,\infty)\mid|f(x)|>\frac{1}{x\ln x}\}$.
Observe that $\int_{A}\frac{1}{x\ln x}dx\leq\int_{A}|f(x)|dx\leq\int|f|<\infty$.
Note that for each $n\geq2$, $[n,\infty)\not\subseteq A$. (For,
if there exists $n\geq2$ such that $[n,\infty)\subseteq A$, then
$\infty=\ln\ln x\mid_{n}^{\infty}=\int_{[n,\infty)}\frac{1}{x\ln x}dx\leq\int_{A}\frac{1}{x\ln x}dx<\infty$,
which is a contradiction). Let $B_{n}=[n,\infty)\setminus A\neq\emptyset$.
Consider the family of non-empty sets $\{B_{n}\mid n=2,3,\ldots\}$,
indexed by $\{2,3,\ldots\}$. By the Axiom of Choice, there exists
a choice function $x:\{2,3,\ldots\}\rightarrow\cup_{n}B_{n}$ such
that $x(n)\in B_{n}$ for each $n$. Denote $x_{n}=x(n)$. Note that
$x_{n}\geq n\Rightarrow x_{n}\rightarrow\infty$. $x_{n}\notin A\Rightarrow|f(x_{n})|\leq\frac{1}{x_{n}\ln x_{n}}$.
Hence, $x_{n}|f(x_{n})|\leq\frac{1}{\ln x_{n}}\rightarrow0$ as $n\rightarrow\infty$.
