I'm working with sets of probability measures of the form: $$ \mathscr{P}_p(\Omega) := \left\{\mu \in \mathscr P: \int_\Omega |x|^p d\mu < + \infty\right\} $$ and I'm trying to show that if $p<q$, then $\mathscr P_p(\Omega)\subset \mathscr P_q(\Omega)$
My immediate thought was to try something like Holder's inequality starting with $$ \int|x|^qd\mu = \| |x|^q\cdot 1\|_{L^1(\Omega, \mu)} $$ and trying to find $\|x\|_p$ in some form as an upper bound, but the conjugate exponents don't seem to work out since $p<q$.
My other idea was to consider the largest natural number $m\ge2$ such that $q-mp\ge 0$, and split the integral $$ \int|x|^qd\mu = \int |x|^{mp}|x|^{q-mp}d\mu $$ and try to use a generlized Holder inequality and the fact that $q-mp < p$ since $m$ is maximal. The issue I seemed to run into again was that the conjugate exponent would bring me back to the $q$-norm, which certainly isn't what I want.
I considered Jensen's inequality (which I don't know very well), but I don't think it works out since $t\mapsto t^{p/q}$ isn't convex.
I'd like to still solve this somewhat on my own, so if you can just nudge me in the right direction or critique my ideas, I'd greatly appreciate that. Thank you!
Counter-example: On $(0,1)$ with Lebesgue measure let $x(t)=\frac 1 {\sqrt t}$. Take $p=1, q=2$.
What is true is $\mathscr P_q(\Omega)\subset \mathscr P_p(\Omega)$ and this follows by Hölder's inequality with exponents $\frac q p$ and $\frac q {q-p}$