Please, help me find connections between two types of convergence:
Let $\{X_n\}_{n\ge1}: (\Omega,F,P) \rightarrow (\mathbb{R},Bor)$ be a sequence of r.v., there are two convergences:
1) $X_n \rightarrow X \hspace{0.2cm}(sLip)$, i.e. $\sum_{n\ge1} E|f(X_n) - f(X)| < \infty \hspace{0.2cm}\forall f \in Lip$ and bounded
2) $X_n \rightarrow X \hspace{0.2cm}(c.c.)$, i.e. $\sum_{n\ge1} P(|X_n - X|\le\epsilon) = \infty \hspace{0.2cm}\forall \epsilon > 0$
I know several things about other type of "complete convergence" ($X_n \rightarrow X \hspace{0.2cm}(c.c.)$, i.e. $\sum_{n\ge1} P(|X_n - X|\le\epsilon) < \infty \hspace{0.2cm}\forall \epsilon > 0$) and it's connection with "strong $L^p$" convergence ($X_n \rightarrow X \hspace{0.2cm}(s.-L^p)$, i.e. $\sum_{n\ge1} E(|X_n - X|^p) < \infty$).
Also, I now about the second Borell-Cantelli lemma, but it uses the independence of random variables (which we don't have).
And it is easy to prove that $E|f(\xi_n)-f(\xi)| \le L E|\xi_n - \xi| \le L ||\xi_n - \xi||_{\infty}$ for L-Lipschitz and bounded functions.
But I don't know, how can I apply all these facts to the given situation (or maybe there is another way to solve this problem). If you have any ideas (or some articles to recommend), I will be very pleasant.
Complete convergence does not imply strong Lipschitz convergence.
To see this, take $x_n = 1/n$ and $x = 0$. For every $\varepsilon >0$, we have $|x_n - x| \leq \varepsilon$ for $n$ large enough so that $P(|x_n - x| \leq \varepsilon) = 1$ for $n$ large enough. In particular, $$\sum_{n=1}^\infty P(|x_n-x| \leq \varepsilon) = \infty.$$ On the other hand, taking $f = \arctan$ which is Lipschitz and bounded, we have $$\sum_{n=1}^\infty E\left|f(x_n) - f(x)\right| = \sum_{n=1}^\infty \arctan\left(\frac{1}{n}\right) = \infty$$ since $\arctan(1/n) \sim 1/n$. This proves that $x_n \to x$ (c.c.) but $x_n \nrightarrow x$ (sLip).
Here's a proof that 1) implies 2).
Assume 1). Then taking $f = \arctan$, we have $E\sum_{n=1}^\infty |f(X_n) - f(X)| < \infty$. This implies that $\sum_{n=1}^\infty |f(X_n)-f(X)| < \infty$ a.s., thus $f(X_n) \to f(X)$ a.s. Applying the continuous function $\tan$, it follows that $X_n \to X$ a.s. This shows that (sLip convergence) $\Rightarrow$ (a.s. convergence).
On the other hand, (a.s. convergence) $\Rightarrow$ (c.c. convergence). By contradiction, if $X_n \nrightarrow X$ (c.c.) then there exists $\varepsilon >0$ such that $\sum_{n=1}^\infty P(|X_n-X|\leq \varepsilon) < \infty$. By the Borel-Cantelli lemma, we get that $P(\limsup_n \lbrace |X_n - X| \leq \varepsilon\rbrace ) = 0$, i.e. a.s. there exists $N = N(\omega)$ such that $|X_n - X| > \varepsilon$ for every $n \geq N$. This implies that $X_n \nrightarrow X$ a.s.