Construct a martingale with a given distribution?

Question

Given a random variable Y, is it possible to construct a martingale M such that $$M_1 \stackrel{D}{=} Y$$

I'm not sure how to go about proving that such an M exists under such general conditions, but I feel like this is definitely a standard question...

Answer 1

If random variable $Y$ takes values in $\mathbb R$, you can find a measurable function $f: \mathbb R\to \mathbb R$, such that $f(N)$ has the same distribution as $Y$, where $N$ is a standard Gaussian random variable.

Now, for a Brownian motion $B_t$, define $$ M_t = E(f(B_1)|\mathcal F^B_t)$$