If $\mathbb Q << \mathbb P$ and $E_{\mathbb P}[\vert X_{n}-X\vert ]\xrightarrow{n \to \infty} 0$ then it is not true for $\mathbb Q$

Question

In the solution to justify why $E_{\mathbb Q}[\vert X_{n}-X\vert ]\xrightarrow{n \to \infty} 0$ does not necessarily hold, given $\mathbb Q << \mathbb P$ and $E_{\mathbb P}[\vert X_{n}-X\vert ]\xrightarrow{n \to \infty} 0$, it is stated that:

Defining $Y_{n}:=\frac{dQ}{dP}\vert X_{n}-X\vert$ and assuming that $X_{n}\to X, \mathbb P-$a.s.

Note $E_{\mathbb Q}[\vert X_{n}-X\vert]=E_{\mathbb P}[\frac{dQ}{dP}\vert X_{n}-X\vert]$. Then:

$0=E_{\mathbb P}[\liminf\limits_{n \to \infty}Y_{n} ]\leq \liminf\limits_{n \to \infty}E_{\mathbb P}[Y_{n}]$ by Fatou's Lemma, and this is used to justify why it does not necessarily hold even if $X_{n}\to X, \mathbb P-$a.s.

But I believe it does by simply using $\limsup\limits_{n \to \infty}$

$0=E_{\mathbb P}[\limsup\limits_{n \to \infty}Y_{n} ]\geq \limsup\limits_{n \to \infty}E_{\mathbb P}[Y_{n}]$(*)

and since it always holds that:

$\limsup\limits_{n \to \infty}\geq \liminf\limits_{n \to \infty}$

and by (*) we also have that $\limsup\limits_{n \to \infty}\leq 0\leq \liminf\limits_{n \to \infty}$

so surely $E_{\mathbb Q}[\vert X_{n}-X\vert ]\xrightarrow{n \to \infty} 0$ holds using Fatou's Lemma twice?

Answer 1

Since $(Y_n)$ may not be uniformly upper bounded (i.e. $(-Y_n)$ may not be uniformly lower bounded), your statement that $$ E\left[\limsup_{n\to\infty} Y_n\right]\ge \limsup_{n\to\infty} E[Y_n] $$ is not deduced from Fatou's lemma. Note that even if $E_{\Bbb P}[|X_n - X|]\to 0$and $\Bbb Q \ll \Bbb P$, it can happen that $E_{\Bbb Q}[|X_n-X|] =\infty$ so $E_{\Bbb Q}[|X_n-X|]\to 0$ does not necessarily hold.

Answer 2

One can construct a counterexample such as Simon suggested, e.g. $$\Omega = (0,1),\ \mathbb{P}(dx) = dx,\ \mathbb{Q}(dx) = \frac{dx}{2\sqrt{x}},\ X_n(x) = n^{3/4}\ 1_{(0,1/n)}(x).$$ Then $$\mathbb{E_P}|X_n| = \frac{n^{3/4}}n\rightarrow 0$$ and $$\Bbb{E_Q}|X_n| = \int_0^{1/n}n^{3/4}\ \frac{dx}{2\sqrt{x}} = n^{3/4}\sqrt{\frac1n} = n^{1/4} \rightarrow \infty.$$