I have started with master course in probability and confused with Skorokhod's theorem.

Let us start with some sequence $X_{n}$ of, say, positive random variables, taking values from the same probability space. Assume, that $X_{n}$ converges in distribution to random variable $X$. Next, let us consider some random positive sequence $Y_{n}$, defined on the same probability space as $X_{n}$ and assume that there exists random $n_{0}$ such that for all $n > n_{0}$ we have almost surely $$ Y_{n} \leq X_{n}. $$

Next, since we have a weak convergence for $X_{n}$ and the real line is separable space, then we can construct a sequence $\tilde{X}_{n}$, which has the same distributions as $X_{n}$ and converges a.s. to some $\tilde{X}$ with the same distribution as $X$.

Questions I am faced with:

Does this make sense to write $$ \limsup_{n} Y_{n} \leq X $$ if all the sequence is defined on the same probability space?

Also, is the following

$$ \limsup_{n} Y_{n} \leq \tilde{X} $$

and $$ E[\limsup_{n} Y_{n}] \leq E[\tilde{X}] $$



Your conjecture is not true.The counterexmaple is in following:
By taking $X_n$ to be a sequence of iid random variable whose law is $\mathcal{U}([0,1])$. $Y_n$ is chosen to be equal to $X_n$, that is $Y_n:= X_n$.
So on one hand you see that $$X_n \xrightarrow{ \text{weakly}} \mathcal{U}[0,1]$$ While $$\limsup Y_n = 1 \quad a.e$$


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