Connection between the cumulative distribution function and $p$-integrability

Question

Let $p \geq 1$ and $l > 0$. Suppose that $X$ is a non-negative random variable with a CDF satisfying for some constant $c > 0$, $\lim_{x\to+\infty}(x^l(1 - F_X(x))) = c$.

I have to show that a non-negative random variable $X$ is $p$-integrable iff $p < l$. I already know that the expected value of a non-negative r.v. can be expressed as $E(X) = \int_0^\infty(1 - F_X(x))dx$, and for $X^p$, $E(X^p) = \int_0^\infty px^{p-1}(1 - F_X(x))dx$.

I'm not looking for a complete answer, but some hints would be nice as currently I don't have any clue on how to start the proof. To be specific, I don't know how to use the fact that $\lim_{x\to+\infty}(x^l(1 - F_X(x))) = c$ with the CDF form of the moment $p$ of $X$.

Answer 1

Choose $\epsilon>0$, so that $\epsilon < c$. There exists $\delta$ so that $$ x>\delta \quad \implies \quad c-\epsilon < x^\ell (1-F(x))< c+\epsilon \tag1 $$ Now, write $$ E[X^p]=\underbrace{\int_0^\delta px^{p-1}(1-F(x))\,dx}_{\text {finite}}+\int_\delta^\infty px^{p-1} (1-F(x))\,dx $$ The first integral is always finite, so we focus on the second. Using $(1)$, you can show the second integral is finite when $p<\ell$ and infinite when $p\ge \ell$.

Further details:

When $p<\ell$, you show the second integral is finite by using$$\int_\delta^\infty px^{p-1} (1-F(x))\,dx<\int_\delta^\infty px^{p-1} x^{-\ell} (c+\epsilon)\,dx$$ When $p\ge \ell$, you can show the second integral is infinite by using$$\int_\delta^\infty px^{p-1} (1-F(x))\,dx>\int_\delta^\infty px^{p-1} x^{-\ell} (c-\epsilon)\,dx$$