= Strict stability implies self-decomposability
{title2=$R_c\overset d=(1-c^\alpha)^{1/\alpha}X'$}
Let $X$ have a <strictly stable distribution> of index $\alpha$, and let $X'$ be an independent copy. For integers $1\leq k<n$, split a sum of $n$ independent copies into blocks of sizes $k$ and $n-k$. Strict stability of each block and of the full sum gives
$$
(k/n)^{1/\alpha}X+(1-k/n)^{1/\alpha}X'\overset d=X.
$$
For any $0<c<1$, choose rationals $k_m/n_m\to c^\alpha$. On the fixed <probability space> carrying $X,X'$, the left side converges <almost surely> to $cX+(1-c^\alpha)^{1/\alpha}X'$. Thus it converges in <distribution>; since its <probability distribution> is constant, the limit also has the law of $X$. This proves a <self-decomposable distribution> with the displayed remainder. The proof does not require any finite <moments>.
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