Decompose integrable probability measure into Bernoulli distribution

Question

Given a probability measure $P$ on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ such that a random variable with law $P$ has finite expectation and the expectation equals to 0. Let $\mu_{a, b}$ be the law of Bernoulli distribution at $a$ and $b$ where $a \neq b$ and $a, b\in \mathbb{R}$ such that the expectation is 0. Show that there exists $\nu$ on $(\mathbb{R}^2, \mathcal{B}(\mathbb{R}^2))$, $\nu(\mathbb{R}^2) = 1$ such that $$P(S) = \int_{\mathbb{R}^2} \mu_{a, b}(S)d\nu$$ for all $S\subset\mathbb{R}$. In other words, show that the set of law of integrable random variables is equal to the following set: $$conv\{\mu_{a, b}: a\neq b, a, b\in \mathbb{R}\}$$. Intuitively, this is basically saying we can decompose a random variable with mean 0 into Bernoulli random variables. There isn't a theorem that comes into my mind that I can use for such decomposition. Is there any suggestions for how to start with?