Question regarding convex combination of probability measures

I want to further understand a topic discussed in this post: Intuition of convex combination of probability measures. In a class material I have the following statement:

If $$\nu_1$$ and $$\nu_2$$ are probability distributions and $$\alpha\in[0,1]$$, then a random variable $$X$$ distribution $$\alpha\nu_1 + (1-\alpha)\nu_2$$ iff there exists a random variable $$Y\sim Ber(\alpha)$$ such that $$X$$ has conditional distribution $$\nu_1$$ given $$Y=1$$ and conditional distribution $$\nu_2$$ given $$Y=0$$.

The backward implication is fairly straight forward, as highlighted in the mentioned post: $$P(X\in A) = P(X\in A\mid Y=1)P(Y=1) + P(X\in A\mid Y=0)P(Y=0) = (\alpha\nu_1 + (1-\alpha)\nu_2)(A).$$

I am stuck with the forward implication. It is clear to me that the choice of the variable $$Y$$ is essential (it can't just be any Bernoulli with parameter $$\alpha$$), but I can't figure out how to choose it.

Let's illustrate this with the two distributions having densities $$f_1(x)$$ and $$f_2(x)$$ with the density for the mixture distribution being $$\alpha f_1(x) + (1-\alpha)f_2(x)$$.
Then you can say $$\mathbb P(Y=1 \mid X =x) = \frac{\alpha f_1(x) }{\alpha f_1(x) + (1-\alpha)f_2(x)}$$ which will give you the marginal distribution for $$Y$$ of $$\mathbb P(Y=1)=\alpha$$ as desired and you can deduce the conditional densities $$f_{X\mid Y=1}(x)=f_1(x)$$ and $$f_{X\mid Y=0}(x)=f_2(x)$$.
For more general distributions, you need to translate the conditional probability for $$Y$$ into the limit of the ratio of probabilities in a small neighbourhood around $$x$$ and get the more general result, but it is essentially the same.