Let $X$ be a random variable. Show that $x P(|X|>x) \to 0$ as $x \to \infty$ implies $E[|X|^{1-\epsilon}] < \infty$ for any $0<\epsilon<1$.
My attempt:
We know that if $Y$ is a positive random variable, then by applying Fubini's theorem we have $E[Y^p] = \int_{0}^\infty py^{p-1}P(Y>y)dy$.
Therefore, by using the equation above, I said: $$E[|X|^{1-\epsilon}] = \int_{0}^\infty (1-\epsilon)x^{-\epsilon}P(|X|>x)dx$$
However, this is where I am stuck. We are given that $xP(|X|>x) \to 0$, so I've tried multiplying both sides of the equation by $x^{1+\epsilon}$. By doing so, I get: $$ x^{1+\epsilon}E[|X|^{1-\epsilon}] = (1-\epsilon)\int_{0}^\infty xP(|X|>x)dx $$ As I take $x \to \infty$, the RHS goes to zero (by assumption) but I believe the LHS goes to $\infty$ since $\epsilon>0$. There seems to be a contradiction
Observe that $E[|X|^{1-\epsilon}] = \int_{0}^\infty (1-\epsilon)x^{-\epsilon}P(|X|>x)dx= (1-\epsilon) \int_{0}^\infty x^{-\epsilon-1}[P(|X|>x)x]dx$.
In $0$ you don't have problem in order to conclude the integral is finite. For $x \to \infty$ observe that $\frac{1}{x^{1+\varepsilon}}$ is integrable and $P(|X|>x)x \leq 1$ definitely because $x P(|X|>x) \to 0$. In fact, $x P(|X|>x) \to 0$ implies that for any $\varepsilon > 0$ (Thus also for $\varepsilon=1$) there exists $M(\varepsilon)$ such that for $x \geq M(\varepsilon)$ we have $P(|X|>x)x \leq \varepsilon$.