It is known that a random variable $X$ which is symmetric about $0$ (i.e $X$ and $-X$ have the same distribution) must have all its odd moments (when they exist!) equal to zero. The converse is a strong temptation which has been around for perhaps a hundred years.

Question. Suppose $\mathbb E[X^n] = 0$ for all odd $n$. Is it true that $X$ is symmetric ?

This question was solved Churchill (1946). In fact, he proved something much stronger

Theorem. Let $X$ be a random variable and let $(a_m)_{m \in \mathbb N}$ be a sequence of real numbers. Then for every $\epsilon > 0$, the exists a random variable $Y$ such that (1) $\mathbb E[Y^{2m+1}] = a_m$ for all $m \in \mathbb N$, and (2) The Kolmogorov distance between $X$ and $Y$ is at most $\epsilon$.

Of course this theorem immediately implies a negative answer to the above question.

The proof given in the paper is constructive, but somewhat mysterious.

Question. Is there simple / modern way to prove the above statement using functional-analytic tools ?

After all, the above theory simply says (roughly) that the set of random variables of with odd-moments given by the sequence $(a_m)$ is dense in the space space of random variables equipped with Komolgorov distance.

Note. The expected advantage of a general functional-analytic solution is that it would perhaps extend to even more general constraints those prescribed odd moments.