Does $\operatorname EX_n\to0$ as $n\to\infty$?

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Suppose that $X_1,X_2,\ldots$ are non-negative random variables defined on a probability space $(\Omega,\mathcal F,P)$ with $\operatorname E|X_n|^p<\infty$ with some $p>2$ for each $n\ge1$. Suppose that there exists $\Omega_0\subset\Omega$ such that $P(\Omega_0)=1$ and for each $\omega\in\Omega_0$ there exists $N(\omega)\ge1$ such that $X_n(\omega)=0$ when $n\ge N(\omega)$. In other words, the sequence $\{X_n(\omega):n\ge1\}$ is eventually equal to $0$ for each $\omega\in\Omega_0$.

Can we deduce that $\operatorname EX_n\to0$ as $n\to\infty$?

The first thing that comes to my mind is the monotone convergence theorem. For each $\omega\in\Omega_0$, $0=X_n(\omega)\le X_{n+1}(\omega)=0$ when $n\ge N(\omega)$. If we want to apply the dominated convergence theorem, this inequality has to hold for all $n\ge1$ and all $\omega\in\Omega_0$. We could drop the first $N(\omega)-1$ terms for each $\omega\in\Omega_0$, but $N(\omega)$ depends on $\omega$ so it is not necessarily possible to have $X_n(\omega)\le X_{n+1}(\omega)$ for each sufficiently large $n$.

Any help is much appreciated!

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No. Example: $X_n =nI_{(0,1/n)}$ on the space $(0,1)$ with Lebesgue measure.