We know that $B_t \sim N(0,t)$ has the same distribution of $\sqrt tB_1 \sim \sqrt tN(0,1)=N(0,t)$
1) why it is wrong do the following: $E[B_tB_s] = \sqrt t \sqrt sE[B_1B_1]$. Instead it hold that $E[B_t-B_s]=E[\sqrt tB_1-\sqrt sB_1]$ , but I guess it is only a coincidence here.
2) often we sum and subtract a brownian motion for bringing us back to an increment and exploiting the independence. Summing and subtracting a r.v.s does not change the result, indeed $+B_s-B_s=0$. However this result hold because we pick up this two Brownian motion from the same probability space. If they come from a different space the result does not hold. Until here it is right? (please give a feedback also on this)
But what does it mean taking two Brownian motion from a different probability space? The answer should be that for a different $\omega$, for a fixed $t$, the r.v. $B_t$ assume a different value w.r.t. another Brownian motion, say $W_t$. Could someone provide a more concrete example of two probability spaces where I can define two Brownian motion?
Brownian Motion has only independent increases ( that is $B_{t_1} , B_{t_2} - B_{ t_1} , ... B_{t_k} -B_{t_{k-1}} $ are independent for $0\le t_1 < t_2 < ... <t_k$. So writing $B_t = \sqrt tB_1$ isn't the best choice, because then $B_{t_1} = \sqrt t_1B_1 $ and $B_{t_2} - B_{t_1} = (\sqrt{t_2} - \sqrt{t_1})B_1 $ are clearly dependent.
So you can $\mathbb E[B_t - B_s] = \mathbb E[B_t] - \mathbb E[B_s] = 0-0 = 0$, since $B_x \sim \mathcal N(0,x)$
But assuming $t\ge s$ , $\mathbb E[B_sB_t] = \mathbb E[B_sB_t - B_sB_s +B_sB_s] = \mathbb E[B_s^2] + \mathbb E[B_s(B_t-B_s)]$. The first one is just the second moment of $\mathcal N(0,s)$, and the second, due to independence and symetry of $\mathcal N(0,s)$ is $0$. When you wanted to plug $B_t = \sqrt{t}B_1, B_s = \sqrt{s}B_1$ you will lose your independence of increments, so it isn't good.
As for the last question, you know, there are many different random variables having the same distribution. Say you have Brownian Motion on some probability space $(\Omega, \mathcal F, \mathbb P)$, that is $B:\Omega \to C[0,1]$, $B = (B_t)_{t \in [0,1]}$ where you have a set $A \in \mathcal F$ such that $\mathbb P(A) = 0$. Then you can do the following: Define $W:\Omega \to C[0,1]$, $W = (W_t)_{t \in [0,1]}$ such that $W(\omega) = B(\omega)$ for every $\omega \in \Omega , \omega \notin A$ and $W(\omega) = c_\omega$ for $\omega \in A$ $($By $c_\omega$ I mean the $c_\omega-$ constant function (dependent of $\omega$) $c_\omega(t) = c_\omega$, for any $t \in [0,1])$
Then clearly both $W,B$ have the same distribution, since for any $E \in \mathcal B(C[0,1])$ (borel set), since we have $(\mu_W -$distribution of $W$ , $ \mu_B -$ distribution of $B)$
$\mu_{W}(E) = \mathbb P( \{ \omega \in \Omega : W(\omega) \in E \} = \mathbb P( \{ \omega \in \Omega : W(\omega) \in E, W(\omega) \notin A \}) = \mathbb P(\{\omega \in \Omega: B(\omega) \in E, B(\omega) \notin A \}) = \mathbb P(\{ \omega \in \Omega: B(\omega) \in E\} = \mu_B(E)$
But when we take $c_\omega \neq B(\omega)(0)$ (value of $B(\omega)$ at $t=0$ (cause $B$ is random variable to set $C[0,1]$, then for any $\omega \in A$ we have $W(\omega) \neq B(\omega)$, so they are different, but still both has the same distribution of BrownianMotion.
If you want different probability spaces, note that you can just define $W_A : \Omega_A \to C[0,1]$, where $\Omega_A = \{ \omega \in \Omega : \omega \notin A \}$ and $W_A(\omega) = W(\omega)$ for any $\omega \in \Omega_A$
Note I gave you example of both: first -(the same probability space, different R.V, but still both brownian motion) and the second (different probability space, still brownian motion).