# understanding Skorokhod's theorem and convergences

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

correct?

## 1 Answer

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 $$\limsup Y_n = 1 \quad a.e$$