Skorohod Representation Theorem Application I'm reading through the Skorohod representation theorem and finding this proof a little bit difficult to understand. My understanding of the proof in bullet point form is as follows:


*

*Define a new random variable Y on the same probability space as X.

*$Y_{n}$ has the same probability distribution as $X_{n}$ for all
positive n.

*Y has the same distribution as X.

*$Y_{n} \to Y$ as $n \to \infty$ with probability 1.

*Since g is a continuous func, $g(Y_{n}) \to g(Y)$ as $n \to \infty$
with probability 1.

*Thus $g(Y_{n}) \overset{d}\to g(Y)$ as $n \to \infty$.

*Lastly, we argue that $g(Y_{n})$ has the same distribution as
$g(X_{n})$ and $g(Y)$ has the same distribution as $g(X)$ so we are
done.
My questions are as follows: how do we get from step 2 to step 3? Is step 4 a direct result of using the Skorohod representation theorem? Why does $g(Y_{n}) \to g(Y)$ simply because g is a continuous function?


Step 3 doesn't follow from step 2. Assertions 2,3,4 are directly a result of the Skorohod representation theorem; they correspond to the conclusions a,b,c in the Skorohod theorem that your link points to.
As for the claim that $g(Y_n)\to g(Y)$, this follows from continuity of $g$: If $x_n$ is a sequence of real numbers that converge to $x$, then $g(x_n)\to g(x)$ whenever $x$ is a continuity point for the function $g$. Applying this to $Y_n$ and $Y$, this means
$$
\{Y_n\to Y\}\subset \{g(Y_n)\to g(Y)\}$$
since, by assumption, $g$ is continuous at every point. But the event on the LHS has probability 1, therefore the event on the RHS also has probability 1.
