Could someone please explain how this holds:
$\displaystyle \int_{\mathbb{R^n}} f d\mu = \int_{\Omega}f(Y_n)d\mathbb{P}$
Does it use the following proposition?
Furthermore how does $\mathbb{E}(f(X_n)) \rightarrow \mathbb{E}(f(X))$ hold? It is not clear to me how proposition 3.4 is used.
The step $$\int_{\mathbb R}f ~d\mu_n=\int_{\Omega}f(Y_n)~d\mathbb P$$ in Theorem 3.9 is indeed a consequence of Proposition 3.4.
Consider the measure spaces $(\Omega,\mathcal A,\mathbb P)$ and $(\mathbb R,\mathscr B(\mathbb R),\mu_n)$ (where $\mathscr B(\mathbb R)$ denotes the Borel $\sigma$-algebra). Since $Y_n$ is a random variable on the probability space $\Omega$, it is by definition a measurable map from $\Omega$ to $\mathbb R$. Furthermore, we notice that for every measurable set $A\subset\mathscr B(\mathbb R)$, we have that $$\mu_n(A)=\mathbb P[Y_n\in A]=\mathbb P(Y_n^{-1}(A))=(\mathbb P\circ Y_n^{-1})(A),$$ hence $\mu_n=\mathbb P\circ Y_n^{-1}$. Consequently, by applying Proposition 3.4 directly, we obtain that for any measurable function $f$, we have that $$\int_{\mathbb R}f~d\mu_n=\int_{\Omega}f\circ Y_n~d\mathbb P.$$ Given that every continuous function is measurable, this gives you the equality you wanted.
As for $\mathbb E(f(X_n))\to\mathbb E(f(X))$, as mentioned in the statement of Theorem 3.9, this is a direct consequence of Proposition 3.4. Indeed, if we generalize the steps we did in the previous paragraph to an arbitrary random variable $X$ with distribution $\mu$, then for any measurable function $f$, we know that $$\mathbb E(f(X))=\int_{\Omega}f(X)~d\mathbb P$$ by definition of the expected value of the random variable $f(X)$ (indeed, a composition of measurable maps is measurable, so $f(X)$ is a random variable), and thus $$\mathbb E(f(X))=\int_{\Omega}f(X)~d\mathbb P=\int_{\mathbb R}f~d\mu.$$ Therefore, saying that $\mu_n\Rightarrow\mu$ is really exactly the same as saying that $\mathbb E(f(X_n))\to\mathbb E(f(X))$ for every $f$ continuous and bounded, where $X_n\sim\mu_n$ and $X\sim\mu$.