Consider a measure space $(X, \mathcal{F}, \mu)$ and let $f \in L^1(X, \mathcal{F}, \mu)$ with $f(x) >0$ for all $x \in X$ be a probability density function. As discussed in this question, $f$ need not be in $L^2(X, \mathcal{F}, \mu)$. Moreover, $f$ can be continuous and differentiable and still not be square-integrable.
My question is if there are "simple" assumptions one can make about $f$ such that it lies in $L^2(X, \mathcal{F}, \mu)$.