Almost sure convergence implies convegence in distribution - proof using monotone convergence I'm trying to understand the following proof of the statement : "Almost sure convergence implies convegence in distribution"
The definition of convergence in distribution is given as follows : 
$X_n$ converges in distribution to $X$ if and only if for all bounded real function $f$ we have :
$$\lim_{n \rightarrow +\infty  }E\left[f(X_n)\right]=E\left[f(X)\right]$$
The proof goes like this :
If $X_n$ converges almost surely to $X$ then $f(X_n)$ converges almost surely to $f(X)$. Now using the dominated convergence theorem which is :
$$\lim_{n \rightarrow +\infty  } \int f_n d\mu = \int f d\mu $$
we get : 
$$\lim_{n \rightarrow +\infty  }E[f(X_n)]=E[f(X)].$$
My question is this : How using the dominated convergence theorem gets us from : 
$$f(X_n)\rightarrow^{a.s.} f(X)$$ to $$\lim_{n \rightarrow +\infty  }E[f(X_n)]=E[f(X)]$$
given that the almost sure convergence is given by : $P\left[\lim_{n \rightarrow + \infty} X_n = X\right] = 1$?
One of my attempts is to write the convergence in distribution in the form of integrals like this : 
$$\lim_{n \rightarrow +\infty  }E[f(X_n)]=E[f(X)]$$ 
is equivalent to :
$$\lim_{n \rightarrow +\infty  }\int f(y) \phi_n(y) dy=\int f(y) \phi(y) dy$$ 
with $\phi$ the density of $X$ and $\phi_n$ the density of $X_n$. But this is a little different from the monotone convergence theorem result. In the dominated convergence theorem we have the same measure, $\mu$, but writing the expectations gives us two different measures, which are $\phi_n dy$ and $\phi dy$, respectively the cumulative distributions for $X_n$ and $X$
Any help please? I appreciate if you can tell me why my attempt is not leading anywhere and at the same time give your own proof. Thank you!
 A: In order to apply dominated convergence, write the expectations as integrals over the probability space $(\Omega,{\cal F},P)$, not the real line:
$$E(f(X_n))=\int_\Omega f(X_n(\omega))\,P(d\omega)\to\int_\Omega f(X(\omega))\,P(d\omega)=E(f(X)).$$
A: As the comment by Danielsen points out, the most common definition of the convergence in distribution is $$\lim_{n \rightarrow +\infty  }\mathbb E\left[f\left(X_n\right)\right]=\mathbb E\left[f(X)\right]$$
for each continuous and bounded function $f$.
If $f$ is such a function and $X_n\to X$ almost surely, then $X_n(\omega)\to X(\omega)$ for each $\omega\in\Omega\setminus N$, where $N$ is a measurable set of probability zero. Since $f$ is continuous, we have $\left(f(X_n)\right)(\omega)\to\left(f(X)\right)(\omega)$ for each $\omega\in \Omega\setminus N$, hence $f(X_n)\to f(X)$ almost surely. Since the function $f$ is supposed to be bounded, the sequence $\left(f\left(X_n\right)\right)_{n\geqslant 1}$ is dominated by a constant hence integrable function. 
