Does $\mathbb{E}[X^6] < \infty$ imply $\mathbb{E}[X^4] < \infty$? 
Let $X$ be a random variable.
  Does $\mathbb{E}[X^6] < \infty$ imply $\mathbb{E}[X^4] < \infty$?

My tries


*

*I know this isn't rigorous but I thought that when there was a counterexample, the rv in question would have density so I could write
$$
\mathbb{E}[X^k]
= \int_{\mathbb{R}} x^k f(x) \ \text{d} x,
$$
where $f$ is the PDF of $X$. 
I thought that if I had a rv with density $f(x) := x^{-4}$ this would yield $\mathbb{E}[X^4] = 1$ and $\mathbb{E}[X^6] = \infty$ but as $\int_{\mathbb{R}} x^{-4} \ \text{d} x = \infty$, $f$ isn't a PDF, so I don't know where to continue from there. 

*Jensens inequality gives $\mathbb{E}[X^k] \ge \mathbb{E}[X]^k$ from $k \in \mathbb{N}$ but if $\mathbb{E}[X] \in (0,1)$ we have $\mathbb{E}[X]^6 < \mathbb{E}[X]^4$ and for $\mathbb{E}[X] > 1$ we have $\mathbb{E}[X]^6 > \mathbb{E}[X]^4$, so I don't know where to continue from there. 

*I wanted to use that $L^q(\Omega) \subset L^p(\Omega)$ for $p \le q$ if $\Omega$ is a finite measure space but I have no reason to believe I can restrict myself to that special case.

Jensen's inequality in fact gives $E[Y^k] \ge E[Y]^k$ for any nonnegative random variable $Y$ and any $k \ge 1$ ($k$ does not have to be an integer).  Now apply this with $Y = X^4$ and $k=6/4$.
