Q:

Show: $X_n\to X\text{ a.s.}\implies X_n\to X\text{ stochastically }\implies X_n\to X\text{ in distribution}$

Let $X_1,X_2,X_3,\ldots$ be random variables on one probability space $(\Omega,\mathcal{A},P)$. Show
    $$
X_n\to X\text{ a.s.}\implies X_n\to X\text{ stochastically }\implies X_n\to X\text{ in distribution}
$$

First of all I give you our definitions of the different meanings of convergence because my experience taught me that there are different words for the same things.
1) Convergence a.s.
A sequence $(f_n)$ of numerical measurable functions converges a.s. if there is a null set $N$ so that for $\omega\notin N$ it is $f(\omega)=\lim_n f_n(\omega)$.
2) Convergence stochastically
A sequence $(f_n)$ of measurable functions converges locally stochastically to the measurable function $f$ if for all $A\in\mathcal{A}$ with $P(A)<\infty$ and all $\varepsilon >0$ it is $\mu(A\cap\left\{\lvert f_n-f\rvert\right\})\to 0$. If this holds for $\Omega$, too, then it is called stochastically convergent.
3) Convergence in distribution
For all bounded continious functions $f$ it is
$$
\lim_n \int f\circ X_n\, dP=\int f\circ X\, dP.
$$

Now to what I've tried to make the proof:
First "$\Rightarrow$": If $X_n\to X$ a.s. this implies that $(X_n)$ converges stochastically (locally!) to $X$. But because here it is $P(\Omega)=1<\infty$ the sequence is not only locally stochastically convergent to $X$ but also stochastically.
Second "$\Rightarrow$":
Let $f$ be bounded and continious. I have to show then
$$
\int f(X_n)\ dP\to\int f(X)\, dP.
$$
Do not know how to show that. Can you give me some help, please?

A:

First implication is correct.
For the second one:
$$\left|\int f\circ X_n\, dP - \int f\circ X\, dP\right| \le 
\int | f\circ X_n\ - f\circ X | \,dP \\=
\int_{A_n} | f\circ X_n\ - f\circ X | \,dP + \int_{\Omega - A_n} | f\circ X_n\ - f\circ X | \,dP
$$
Here, let $\epsilon >0$, $\eta$ such as $|x_1 - x_2| < \eta \Rightarrow 
|f(x_1) - f(x_2)| < \frac\epsilon 2$, $A_n = \{ |X_n - X| < \eta\}$ (you can assume $f$ is uniformly continuous, via for instance approximation of $f$ by such functions).
$$\int_{A_n} | f\circ X_n\ - f\circ X | \,dP \le \frac \epsilon 2$$ $$
\int_{\Omega - A_n} | f\circ X_n\ - f\circ X | \,dP \le 2||f|| P(|X_n - X| \ge \eta)$$
Now if $n$ is big enough, $P(|X_n - X| \ge \eta) \le \frac\epsilon{4||f||}$.
Thi gives $$\int f\circ X_n\, dP \to \int f\circ X\, dP$$ as desired.

