Infinite divisibility of random variable vs. distribution The distribution of any infinitely divisible random variable is itself infinitely divisible. But this link says the converse is not always true. Can you explain?
 A: 
That a random variable is infinitely divisible should mean exactly that its distribution is infinitely divisible.

The question originates from some assertions on the page about infinite divisibility of encyclopediaofmath. There the authors define a Poisson random variable $X$ on some probability space $\Omega$ and they note that, on this specific probability space $\Omega$, one cannot decompose $X$ as $X=X_1+X_2$ for some i.i.d. random variables $X_i$, nor as $X=X_1+\cdots+X_n$ for some i.i.d. random variables $X_i$, for any $n\geqslant2$.
Unsurprisingly, to observe this phenomenon, the authors define $X$ on the smallest possible probability space $(\Omega,\mathcal F,P)$, namely, they consider $\Omega=\{0,1,2,\ldots\}$, $\mathcal F=\mathcal P(\Omega)$, $P(\{k\})=\mathrm e^{-\lambda}\lambda^k/k!$ for every $k\geqslant0$, and $X:\Omega\to\mathbb R$ defined by $X(k)=k$. Then, roughly speaking, there is not enough space in $\Omega$ to define some Poisson $\frac12\lambda$ random variables hence the decomposition $X=X_1+X_2$ is impossible on this probability space.
This kind of conundrum is a reason why one usually says that a random variable $X$ on a probability space $\Omega$ is infinitely divisible if there exists some (possibly much larger) probability space $(\Omega',\mathcal F',P')$ and some random variables $X'$, $X'_1$ and $X'_2$ defined on $\Omega'$ such that $X'=X'_1+X'_2$, $X'_1$ and $X'_2$ are i.i.d., and the distributions of $X$ and $X'$ coincide (and likewise for every $n\geqslant2$ instead of $n=2$). Then a random variable is infinitely divisible if and only if its distribution is infinitely divisible.
Ultimately, all this confirms that infinite divisibility, to be useful, should  concern distributions rather than random variables (and in fact I personally never encountered the notion used specifically for random variables).
Edit: Here is a proof that, if $0\lt\lambda\lt.9$, one cannot construct even a single Poisson$(\frac12\lambda)$ random variable $X_1$ on the minimal probability space used above to define the Poisson$(\lambda)$ random variable $X$. Let $p_n=P(X=n)$ and $q=P(X_1=0)$. Assume that $1-p_0\lt q$, then $[X=0]$ must be sent to $[X_1=0]$. Assume that $1-p_1\lt q$, then $[X=1]$ must be sent to $[X_1=0]$. Assume that $p_0+p_1\gt q$, then there is a contradiction since $[X\leqslant1]$ cannot be sent to $[X_1=0]$. Hence the construction is impossible if $1-\mathrm e^{-\lambda}\lt\mathrm e^{-\lambda/2}$,  $1-\lambda\mathrm e^{-\lambda}\lt\mathrm e^{-\lambda/2}$ and  $(1+\lambda)\mathrm e^{-\lambda}\gt\mathrm e^{-\lambda/2}$. These inequalities hold for every $\lambda$ in $(0,\lambda_*)$ where $\lambda_*$ is the unique positive root of $\mathrm e^{\lambda}-\mathrm e^{\lambda/2}=\lambda$, numerically $\lambda_*=.91$.
The problem of the simultaneous construction of $(X_1,X_2)$ on the minimal probability space used above to define the Poisson$(\lambda)$ random variable $X$ is even simpler since every random variable on this space is a function of $X$. But the only way that some random variables $u_1(X)$ and $u_2(X)$ are independent is that one of them (or both) is constant. This cannot happen if $u_1(X)$ and $u_2(X)$ are Poisson$(\frac12\lambda)$ hence $(X_1,X_2)$ do not exist.
