Kolmogorov 0-1 law with almost sure convergence Recently I've tried to solve next problem: consider the sequence of $\{X_n\}_{n=1}^{\infty}$ of independent random variables and $X_n \rightarrow X$ in probability. Prove that the limit $X$ is degenerate ($\exists c \in R: P(X = c) = 1$). 
The solution is: as we have convergence in probability we can find a subsequence $n_k$ in $X_n$ such that $X_n \rightarrow X$ almost sure. After that, we need use Kolmogorov 0-1 law, but I don't really understand how. 
So, the question is: Consider sequence $X_n$ of independent random variables defined on probability space. We know that $X_n$ converge almost sure to $X$. 
I would like to use Kolmogorov 0-1 law to show that $X$ is degenerate. 
For that, I need to show that $X$ is measurable with respect to tail sigma algebra. Why that's true? 
 A: Let $(\Omega,\mathcal{A},\mathbb{P})$ your probability space and $X_n,X:\Omega \to \mathbb{R}$ your random variables such that $X_n \to X $ a.e . Denote  $\mathcal{C}_{\infty} = \cap_{m=1}^{\infty} \sigma(X_m,X_{m+1},...)$ the tail sigma algebra of $X_n$.  
Now let $A=\{\omega \in \Omega : X_n(\omega) \to X(\omega)\}$ which $\mathbb{P}(A)=1 $ and 
$C=\{\omega \in \Omega : \lim_{n \to \infty}X_n(\omega) \text{ exists \}}$. Now you can show that $A \subseteq C$ and $C\in \mathcal{C}_{\infty}\text{(exercise)}$. So we have that $\mathbb{P}(C)=1.$ 
Note: In general it is not true that $A=C$. 
If $A\neq C$ am not sure if it is possible to formally show that $A \in \mathcal{C}_{\infty}$(am not even sure if this is true) but if $A=C$ you can procceed like this:
Let $b \in \mathbb{R}$ and define $B = C\cap X^{-1}((-\infty,b)).$ 
Fix $m \in \mathbb{N}$ and i'll show that $B \in \sigma(X_m,X_{m+1},...)$ and this will imply that $B \in \mathcal{C}_{\infty}\text{ (why? :P)}.\\$
For this let $\epsilon>0$ and define $B_{n}^{\epsilon} = C\cap X_{n}^{-1}((-\infty,b-\epsilon))$ for $n \geq m.$ Now observe that 
$B_{n}^{\epsilon} \in \mathcal{\sigma}(X_m,X_{m+1},...)$ for all $n\geq m$ and $B=\bigcup_{k=1}^{\infty}\bigcup_{l=m}^{\infty}\bigcap_{n=l}^{\infty}B_{n}^{\frac{1}{k}}\  \text{( exercise)}$.
And this gives us that $B \in \mathcal{\sigma}(X_m,X_{m+1},...).\\$
So we have that $B=C\cap X^{-1}((-\infty,b)) \in \mathcal{C}_{\infty}$ so the set 
$C\cap X^{-1}((-\infty,b])\in \mathcal{C}_{\infty} \ \text{( exercise)}\\$.
So for every $b \in \mathbb{R}$ you have that $\mathbb{P}(C\cap X^{-1}((-\infty,b]))\in \{0,1\}$.
Now let $C_b=C\cap X^{-1}((-\infty,b]).$ Since $C_b \nearrow C$ as $b\nearrow \infty$ we have that $\mathbb{P}(C_b)\nearrow \mathbb{P}(C)=1$.
So there exist $b_0$ such that $\mathbb{P}(C_{b_0})=1$ which means that the set 
$S=\{b\in \mathbb{R} : \mathbb{P}(C_b)=1\}$ is non empty and since $C_b \searrow \emptyset$ when $b\searrow -\infty$ we have that there exist $b_1 $ such that $\mathbb{P}(C_{b_0})=0$ which means that $S$ is bounded above.
Let $s*=\inf(S)$ now for every $n \in \mathbb{N}$ we have $\mathbb{P}(C\cap X^{-1}((-\infty,s^*+\frac{1}{n}])=1$ so 
$\mathbb{P}(C\cap X^{-1}((-\infty,s^*]))=\mathbb{P}(\bigcap_{n=1}^{\infty}C\cap X^{-1}((-\infty,s^*+\frac{1}{n}])=1\text{ (exercise).}$ So $s^*\in S.\\$
On the other hand since $s^*=\inf(S)$ we have that $\mathbb{P}(C\cap X^{-1}((-\infty,s^*-\frac{1}{n}]))=0$ for all $n \in \mathbb{N}$ 
so  $ \mathbb{P}(C\cap X^{-1}((-\infty,s^*)))=\mathbb{P}(\bigcup_{n=1}^{\infty}C \cap X^{-1}((-\infty,s^*-\frac{1}{n}]))=0\text{ (exercise).}$
Summarizing the above we have that $\mathbb{P}((-\infty,s^*))=0$ and $\mathbb{P}((-\infty,s^*])=1 $ which means that $\mathbb{P}(X=s^*)=1.\\$
And eventually we have $\mathbb{P}(X=s^*)=1$ , $ \mathbb{P}(X=s)=0$ for all $s<s^*$ and $ \mathbb{P}(X=s)=1$ for all $s>s^*.$
