Existence of indepedent random variables on $(\mathbb N, \mathcal P(\mathbb N))$ Can we find positive numbers $\alpha_1,\alpha_2,\cdots$ such that $\sum \alpha _{n}=1$
and there exists a sequence of non-constant independent random variables on $(%
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\mathbb{N}
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,\mathcal{P}(%
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),P)$ where $P\{n\}=\alpha _{n}$ for each $n$?
What I know:

*

*There is a choice of $(\alpha_n)$ for which no two non-constant random variables on this space are independent.


*On $(0,1)$ with Lebesgue measure we can construct a sequence of indepedent r.v's with given distributions.
Since we cannot find independent sequences on a finite sample space I am wondering if it can be done on countable sspce, specifically on $(\mathbb N, \mathcal P(\mathbb N))$.
I am wondering if an isomorphism theorem can be used to show that this is not possible.
 A: Suppose $X_1,X_2,...$ was such an at most countable sequence of random variables (i.e. non-constant, $\textbf{OR almost-surely non-constant}$).
For the almost-surely non-constant case, I believe the proof I have written  below goes through without the additional assumption that $P(n) > 0$ for all $n \in \mathbb{N}$, to read the proof of this stronger statement, see the ($\textbf{resp.}$ ...) parenthesized additions.
Then $\bigcup_{j=1}^{\infty} X_j(\mathbb{N})$ is some at most countable subset of $\mathbb{R}$, and in particular is borel isomorphic to $(\mathbb{N},2^{\mathbb{N}})$ or a finite subset of this. (By $\mathbb{N}$ I will always mean the set $\{1,2,3,..... \}$, i.e. $\mathbb{N}_{\geq 1}$).
Thus we can, wlog assume that we have $X_1, X_2, ....$ is an at most countable independent sequence of non-constant random variables ($\textbf{resp.}$ $\textbf{almost-surely non-constant}$ random variables) from $(\mathbb{N}, 2^{\mathbb{N}}, P)$ where $P$ is some probability measure satisfying $P(j) = \alpha_j > 0$ for all $j$, to $(\mathbb{N}, 2^{\mathbb{N}})$ ($\textbf{resp.}$ without any assumptions on $P$ besides that it is a probability measure.)
Then $\sigma(X_j)$ is generated by the countable partition $C_j = \{X_j^{-1} (n)\}_{n=1}^{\infty}$ which is also a $\pi$-system, because pairwise intersections of any two sets in this countable partition are empty. Let us also ignore all $P(X_j^{-1}(n)) = 0$ sets in this partition, which amount to ignoring all $X_j^{-1}(n)= \emptyset$ sets (since these are one and the same if we assume $P$ assigns positive mass to every atom), this amounts to setting $C_j = \{X_j^{-1}(n) \}_{n \in X_j(\mathbb{N})}$ since $P(n) > 0 $ for all $n \in \mathbb{N}$ ($\textbf{resp.}$ $C_j = \{X_j^{-1}(n) \}_{n \in \mathbb{N} ; P(X_j^{-1}(n) \neq 0 }$), we now always mean when we write $C_j$ the partition with all empty sets ($\textbf{resp.}$ $P$-null sets) removed.
Therefore $X_1,X_2,....$ are independent if and only if for any $k \geq 1$ and any selection of events $(A_1,...,A_k) \in C_1 \times ... \times C_k$ , we have that $P(A_1 \cap ... A_k) = P(A_1)...P(A_k)$.
$X_j$ is $\textbf{almost-surely}$ non-constant if and only if there is no event $A_j \in C_j$ occuring with probability one, thus $X_1,X_2,...$ are  are all $\textbf{almost surely}$ non-constant if and only if each $C_j$ contains more than one element.
Next, we have $$P(A_1) = \sum_{(j_2,j_3,...,j_k) \in \mathbb{N}_{\geq 1}^{k-1} ; (A_{j_2},...,A_{j_k}) \in C_2 \times .... \times C_k} P(A_1 \cap A_{j_2} .... \cap A_{j_k}) < P(A_1)P(A_{j_2})...P(A_{j_s}) \sum_{(j_{s'},...,j_k) \in \mathbb{N}_{\geq 1}^{k-(s-1)} ; j_s' \neq j_s ; (A_{j_{s'}},...,A_{j_k}) \in C_{s} \times ... \times C_k} P(A_{j_{s'}} \cap ... \cap A_{j_k}) < P(A_1)P(A_{j_2})...P(A_{j_s}) $$,
For any $\textbf{fixed}$ $j_2,....j_s$ , and any $s \leq k$ such that $(A_{j_2},....,A_{j_s}) \in C_2 \times ... \times C_s$, because for any such $\textbf{fixed}$ $j_2,....j_s$ for which $(A_{j_2},....,A_{j_s}) \in C_2 \times ... \times C_s$, the collection of sets $(A_{j_{s'}} \cap ... A_{j_k} )$ with $(j_{s'},...,j_k) \in \mathbb{N}_{\geq 1}^{k-(s-1)} ; (A_{j_{s'}},...,A_{j_k}) \in C_{s} \times ... \times C_k$ forms a countable partition of $\mathbb{N}$
($\textbf{resp. in the sense of forming a countable partition of a probability one subset of}$ $\mathbb{N}$)
Note here we are not restricting $j_{s'} \neq j_s$, and we have the first strict inequality $<$ because each $C_2,...,C_s$ contains no sets of probability zero by construction $\textbf{and}$ each $C_2,...,C_s$ must contain more than one set as every random variable $X_j$ is assumed non-constant ($\textbf{resp. almost-surely non-constant}$), the final strict inequality holds for the same reason since the sets $(A_{j_{s'}} \cap ... A_{j_k} )$ with $(j_{s'},...,j_k) \in \mathbb{N}_{\geq 1}^{k-(s-1)} ; j_{s'} \neq j_{s}; (A_{j_{s'}},..., A_{j_k}) \in C_{s} \times ... \times C_k$ fall short of being a countable partition of $\mathbb{N}$
($\textbf{resp. fall short of forming a countable partition of a probability one subset of}$ $\mathbb{N}$, i.e. they form a countable partition of a subset of $\mathbb{N}$ with probability strictly less than one), because $A_{j_{s}}$ does NOT intersect any of these sets, and has positive mass, since $A_{j_{s}} \in C_s$, which is a collection of positive mass sets by construction.
In other words, we conclude that $P(A_1) < P(A_1)P(A_{j_2})....P(A_{j_s}) < P(A_1)$ for any $s > 1$ and $(A_{j_2},....,A_{j_s}) \in C_2 \times .... \times C_s$, where the final inequality holds because $P(A_{j_t}) \in (0,1)$ for all $t = 2,...,s$ such that $A_{j_t} \in C_t$ by our construction of the $C_t$'s, which is absurd.
Thus we conclude that there exists no sequence of such independent, non-constant
($\textbf{resp. almost-surely}$ non-constant) random variables, under the assumption that the probability measure on $\mathbb{N}$ has a positive mass on every atom ($\textbf{resp.}$ even without this assumption), in fact, the very same proof seems to imply that there can not even exist, for any configuration of $\alpha_j$, two non-constant independent random variables $X_1, X_2$.
($\textbf{resp. almost-surely}$ non-constant), under the assumption that $P(n) > 0$ for all $n \in \mathbb{N}_{\geq 1}$
($\textbf{resp.}$ even without this assumption).
$\textbf{Edit (Some additional remarks that may be worthy of interest)}$ :
So it seems like the following is true, as you say it is impossible to construct an infinite sequence of non-constant independent random variables on a finite probability space, and we have shown that it is even impossible to construct an infinite sequence of non-constant independent random variables on a countable probability space (assuming our probability measure assigns no atoms zero mass), $\textbf{but we can in fact discard this condition}$, since independence in this setting of the proof I wrote above does not care about sets $P(X_j^{-1}(n)) = 0$, indeed I believe if we discard these sets, the proof will still go through.
$\textbf{I have made some parenthesized edits to the proof to show it still goes through}$, please see all the ($\textbf{resp.}$...) parenthesized addendums.
The only caveat/modification we have to make to establish the stronger result is to insist that the random variables $X_j$ are $\textbf{almost-surely}$ non-constant, that is to say, $P(X_j^{-1}(n)) = 1$ does not hold for any $n \in \mathbb{N}$, which is not exactly the same as saying $X_j$ is $\textbf{literally}$ non-constant, for example we may allow $X_j(\omega) = 1$ occuring with probability $0$, and $X_j(\omega') = 2$ occuring with probability $\frac{1}{2}$, so that $X_j$ is not, as a function, non-constant, but is for all intents and purposes (within the scope of probability theory), $\textbf{almost surely non-constant}$.
So I think we have established the following result:
Up to Borel-isomorphism, the only $\textbf{Polish space}$ or standard Borel space, for which one can always ($\textbf{resp.}$ may be able to in special cases) endow a probability measure, so that there can exist an $\textbf{at most countable}$ sequence ($\textbf{resp.}$ uncountable collection $\{X_{\alpha}\}_{\alpha \in I}$), $X_1, X_2,....$,  of $\textbf{almost-surely}$ non-constant independent random variables, all defined on the same probability space, is $([0,1],\mathcal{B}[0,1])$.
So I believe my proof shows the stronger statement that we can never define two or more independent almost surely non-constant random variables on an at most-countable (standard Borel) probability space (which includes finite spaces, since once we allow $P(n) = 0$, we have in essence reduced to the finite space case as well).
