Existence of infinitely many iid RVs given sets of iid RVs of every finite size Let $X:\Omega \rightarrow \mathbb{R}$ be a random variable on a probability space $\langle \Omega, {\cal F}, P \rangle$.  In the cases I am interested in, we may suppose $X$ has a finite range.  Suppose for all $n \in \mathbb{N}$, there is a set $\{X^n_j\}_{j \leq n}$ of iid (with respect to $P$) random variables of size $n$, where each $X_j^n:\Omega \rightarrow \mathbb{R}$ has the same distribution as $X$.  Does it follow there is an infinite set $\{X_k\}_{k \in \mathbb{N}}$ of iid (again, with respect to $P$) random variables $X_k:\Omega \rightarrow \mathbb{R}$ with the same distribution as $X$?  Note, the question requires each member of $\{X_k\}_{k \in \mathbb{N}}$ to have the same domain as $X$.
 A: The answer to your question is yes, and it follows from this technical lemma:

Lemma. Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space such that there exists a sequence of partitions $\{\mathcal
P_n\}_{n\in\mathbb N}$ where each $\mathcal P_n$ consists of countably
  many disjoint measurable sets whose union is $\Omega$. Suppose that $$
\lim_{n\to\infty}\sup \bigl\{\mathbb P(A)\colon A\in \mathcal
P_n\bigr\}=0.\qquad (\star) $$ Then there exists a random variable
  $Y\colon \Omega\to [0,1]$ whose distribution is uniform on $[0,1]$.

(Terminology note: A set is countable if it injects into $\mathbb N$ - in particular finite sets are countable.)
Before proving the lemma, I'll explain why it answers your question. First, let's 
exclude the case of deterministic $X$ (i.e. $\mathbb P(X=x)=1$ for some $x\in\mathbb R$), when the answer is apparent. Next, note that even if we start with an arbitrary random variable, we can always reduce to the case of countable range by replacing $X$ with $X'=\lfloor kX\rfloor$ for some deterministic $k$ sufficiently large such that $X'$ is not deterministic. Then, by letting $v\in \mathbb R^n$ range over the countable set of values attained by the random vector $(X_1^n,\ldots,X_n^n)$, we obtain a countable measurable partition of $\Omega$ given by
$$
\mathcal P_n=\Bigl(\bigl\{\omega\in\Omega\colon (X_1^n,\ldots,X_n^n)=v\bigr\}\Bigr)_v.
$$
It satisfies the condition $(\star)$, since $\alpha:=\sup_{x\in\mathbb R}\mathbb P(X=x)<1$ by assumption and we have the bound $\mathbb P(A)\leq \alpha^n$ for all $A\in\mathcal P_n$ (by independence). Thus the lemma applies, and we obtain a random variable defined on $\Omega$ which is uniformly distributed on $[0,1]$. Finally, since any iid sequence of random variables can be constructed on the probability space $[0,1]$ (a fundamental and well-known fact - e.g. Theorem 2.19 in Kallenberg's textbook, or prove it yourself using binary expansion) we can compose this standard construction with the measurable function $Y\colon \Omega\to [0,1]$ to obtain the desired iid sequence on $\Omega$.
Proof of the lemma. Let $\mathcal Q_n$ be the partition whose elements are all intersections of elements of $\mathcal P_1,\ldots,\mathcal P_n$. Then $\mathcal Q_n$ is also a countable measurable partition of $\Omega$ and $(\star)$ continues to hold after replacing $\mathcal P_n$ with $\mathcal Q_n$. Moreover, the partitions $\mathcal Q_n$ are nested: for all $m\leq n$, each set in $\mathcal Q_n$ belongs to a unique set in $\mathcal Q_m$. Due to this nesting property, the sets in each partition can be ordered as $\mathcal Q_n=(A_i^n)_{i\in\mathcal I}$ for all $n\in \mathbb N$, such that for all integers $m\leq n$ and all pairs of indices $i\leq j$, if $A_{i'}^m$ contains $A_i^n$ and $A_{j'}^m$ contains $A_j^n$ then $i'\leq j'$.
Using this ordering we construct a sequence of random variables $\{Y_n\}_{n\in\mathbb N}$ as follows. For all $\omega\in\Omega,$ let $i_n(\omega)$ be the index such that $\omega\in A_{i(\omega)}^n$, and let
$$
Y_n(\omega)=\sum_{j\leq i(\omega)}\mathbb P(A_j^n).
$$
Observe that if $0\leq a\leq b\leq 1$ and $a,b$ can be written as partial sums of $\mathbb P(A_j^n)$, then
$$
\mathbb P(a<Y_n\leq b)=b-a.\qquad (\star\star)
$$
Now let $Y(\omega)=\inf_{n\in\mathbb N}Y(\omega)$ and observe (by the nesting property of the partitions) that $(\star\star)$ holds with $Y$ in place of $Y_n$ as well. Sending $n\to\infty$ and using $(\star)$ implies that $a,b$ can be chosen to lie in any non-empty open subinterval of $[0,1]$, and thus $Y$ is uniformly distributed on $[0,1]$. $\square$
