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?
 A: I am assuming we mean Lebesgue integrable. The conclusion is equivalent to the claim that $\liminf_{x \to +\infty} |xf(x)| = 0$. If to the contrary $\liminf_{x \to +\infty} |xf(x)| = \epsilon>0$, we have $|f(x)| > \frac{\epsilon}{2x}$ for $x>N$ for some $N$. Thus derive a contradiction comparing $\int |f|$ and $\int \frac{\epsilon}{2x}$.
It's interesting to ask whether this also holds for functions which are improper Riemann integrable, in the sense that $\lim_{(x,y) \to (\infty, \infty)} \int_{-x}^{y} f(t) \ dt $ exists, where the integrals are Riemann integrals.
It turns out the result doesn't hold in that case. Define a sequence $a_0 = 0$, $a_{j} = 1/j + a_{j-1}$ for $j \geq 1$. Define $f:\mathbb{R} \to \mathbb{R}$ by $f(x) = \mathbf{1}_{x \geq 0}(x) \sum_{j \geq 0} (-1)^j\mathbf{1}_{[a_{j}, a_{j+1})}(x)$. We have $\int_{\mathbb{R}} f = \sum_{j \geq 1} (-1)^{j+1}/j$ in the improper Riemann sense but $f$ only takes on values $-1, 1$ for $x \geq 0$.
A: 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 $n\rightarrow\infty$.
