a.s convergence is "probabilistic" (Breiman, Probability, p. 34) 
Let $\{X_n\}$, $\{X'_n\}$ have the same distribution. Prove that if $X_n \xrightarrow{a.s} X$, there is a random variable $X'$ such that $X'_n \xrightarrow{a.s} X'$.

Some of my classmates gave me a counterexample to me, but they were wrong to me.
I want to prove it by showing that $\{X'_n\}$ is Cauchy. And for this, I need this assertions that I DON'T know whether is true or not :

Conjecture. Suppose $X\overset{\mathcal{D}}{=}X'$ and $Y\overset{\mathcal{D}}{=}Y'$. Is it true that $X-Y\overset{\mathcal{D}}{=}X'-Y'$ ?

 A: In general your conjecture is not true: suppose that $X, Y$ are independent standard normal variables, $X^\prime\sim N(0,1)$ and $Y^\prime = X^\prime$. Then 
$$
X-Y\sim N(0,2)
$$
but
$$
X^\prime - Y^\prime=0.
$$
However, as I said in the comments, if $X,Y$ and $X^\prime, Y^\prime$ are independent, then
$$
F_{X-Y}(t)=\int F_X(y+t)dF_Y(y)=\int F_{X^\prime}(y+t)dF_{Y^\prime}(y)=F_{X^\prime-Y^\prime}(t)
$$
and your conjecture holds.
As for the original problem, first note that since we can always consider the sequence $\{X_i-X\}$, it is enough to consider only the case $X_i\to{0}$ a.s.. Then, indeed, the sequence $X_i$ is Cauchy:
$$
\lim_nP(\sup_{m\geq{n}}|X_m|\geq\varepsilon)=0.
$$
But then, using the triangle inequality, we can prove that $X_i^\prime$ is also Cauchy:
$$
P(\sup_{m\geq{n}}|X_m^\prime-X_n^\prime|\geq\varepsilon)\leq P(\sup_{m\geq{n}}(|X_m^\prime|-|X_n^\prime|)\geq\varepsilon)=P(\sup_m|X_m^\prime|\geq\varepsilon+|X_n^\prime|)\leq P(\sup_m|X_m^\prime|\geq{2\varepsilon},|X_n|\leq{\varepsilon})\leq P(\sup_m|X_m^\prime|\geq{2\varepsilon})=P(\sup_m|X_m|\geq{2\varepsilon})\to0
$$
A: *

*If the distribution of the sequences (as random elements in ${\mathbb R}^{\mathbb N}$) is identical  then YES, a.s. convergence of one is equivalent to a.s. convergene of the other, because convergence of the sequence $(Y_n)$  is the measurable event:
$$\bigcap_{k=1}^\infty \bigcup_{l=1}^{\infty}\bigcap_{m,n\ge l}\{|Y_n-Y_m|<k^{-1}\},$$
which has the same probability under the two seqeunces. 

*If only for every $n$, the distribution of $X_n$ and $X'_n$ is the same then NO. Example: Take $Z_0,Z_1,Z_2$ be IID and nondegenerate. Set $X_n = Z_0$ for all $n$ and set $X'_{2n} =Z_1$ and $X'_{2n+1}=Z_2$. Now, the sequences $(X_n)$ and $(X_n')$ are independent, and $X_n=X_n'$ in distribution for all $n$. The first converges a.s. and the second does not because it does not even converge in probability. 
A: I've found a counterexample for this assertion. What do you think?
Define 
$\left\{\begin{array}{ll}X:([0,1],\mathcal{B}_{[0,1]},m)\rightarrow ([0,1],\mathcal{B}_{[0,1]})\\
X(\omega):=\chi_{[0,\frac12]}(\omega)
\end{array}\right.$ and
$\left\{\begin{array}{ll}Y:([0,1],\mathcal{B}_{[0,1]},m)\rightarrow ([0,1],\mathcal{B}_{[0,1]})\\
Y(\omega):=\chi_{(\frac12,1]}(\omega)
\end{array}\right.$
where $m$ is Lebesgue measure.
Now, clearly, $X\overset{\mathcal{D}}{=}Y$ because they've shared their whole mass equally on $\{0\}$ and $\{1\}$. BUT they have this good property : 
$X(\omega)\neq Y(\omega)\;everywhere$.
Thus, the sequence $<X,X,X,\cdots>$ is same distributed as $<X,Y,X,Y,\cdots>$, but the first one converges $surely$ to $X$ while the second one is not even $Cauchy$ for every $\omega\in[0,1]$.
