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.
 A: Your point $3.$ definitely applies: $\mathbb{E}[X^6]< \infty$ is the same as $X \in L^6(\Omega)$, and hence $X \in L^4(\Omega)$ by point $3.$, which is to say that $\mathbb{E}[X^4]<\infty$.
However, as previously mentioned in the comments, it suffices to see that $X^4 \leq X^6+1$. The inequality can be seen to follow from the fact that for $-1 \leq x \leq 1$, it is true that $x^4 \leq 1$, and for $\left| x \right| \geq 1$, $x^4 \leq x^6$. Hence, $x^4 \leq \max\{x^6,1\}\leq x^6+1$. 
Since $X^6$ is integrable by assumption and $1$ due to the fact the measure space is finite, we have that $X^6+1$ is integrable and therefore $X^4$ is.
A: 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 $k=6/4$.
A: This is an elementary but important consequence of Holder inequality.
I state and prove the following result.
Proposition: Let $(X,\mathcal{M},\mu)$ be a measure space with $\mu(X)<\infty$.
Let $0<a<b<\infty$. For any measurable function $f:X\rightarrow\mathbb{R}$,
we have that $\int|f|^{b}d\mu<\infty\Rightarrow\int|f|^{a}d\mu<\infty$.
Proof: Let $p=\frac{b}{a}\in(1,\infty)$ and let $q\in(1,\infty)$
such that $\frac{1}{p}+\frac{1}{q}=1$. Let $\xi=|f|^{a}$ and $\eta=1$.
By Holder inequality, we have 
$$
\int|\xi\eta|\,d\mu\leq||\xi||_{p}||\eta||_{q}.
$$
Observe that $\int|\xi\eta|\,d\mu =\int|f|^{a}\,d\mu$, $||\xi||_{p}=\sqrt[p]{\int|\xi|^{p}\,d\mu}=\sqrt[p]{\int|f|^{b}\,d\mu}$,
and $||\eta||_{q}=\sqrt[q]{\mu(X)}$. Therefore 
\begin{eqnarray*}
\int|f|^{a}\,d\mu & \leq & \sqrt[p]{\int|f|^{b}\,d\mu}\cdot\sqrt[q]{\mu(X)}<\infty
\end{eqnarray*}
 whenever $\int|f|^{b}\,d\mu<\infty$.
