$L^1(\mathbb{P}) \neq L^1(\mathbb{Q})$ in general? Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and $L^1(\mathbb{P})$ denote the collection of all random variables that are $\mathbb{P}$-integrable.
Now let $\mathbb{Q}$ be an absolutely continuous probability measure w.r.t $\mathbb{P}$ and denote its Radon-Nikodym derivative by $Z$.  Denote the collection of random variables that are $\mathbb{Q}$-integrable by $L^1(\mathbb{Q})$.  We know that $X \in L^1(\mathbb{Q})$ if and only if $ZX \in L^1(\mathbb{P})$.
Is it true that if $Z \notin L^\infty$, then there exists $X \in L^1(\mathbb{P})$ such that $\mathbb{E}[ZX] = \infty$?
I am fairly sure this is true, but I am not sure how to prove it.

By assumption, as $Z \notin L^{\infty}$, we have $$\mathbb{P}(Z > n^2) > 0, \quad n \in \mathbb{N}.$$ Therefore, we may define $$X = \sum_{n=1}^{\infty}\frac{1}{n^2} \frac{\mathbb{1}_{[Z > n^2]}}{\mathbb{P}(Z>n^2)} \in L^1(\mathbb{P}).$$ However, it follows that
\begin{align*}\mathbb{E}[ZX] = \sum_{n=1}^{\infty}\mathbb{E}\left[\frac{1}{n^2} \frac{Z\mathbb{1}_{[Z > n^2]}}{\mathbb{P}(Z>n^2)}\right] \geq \sum_{n=1}^{\infty}\mathbb{E}\left[\frac{1}{n^2} \frac{n^2\mathbb{1}_{[Z > n^2]}}{\mathbb{P}(Z>n^2)}\right] = \sum_{n=1}^{\infty}1 = \infty.\end{align*}
