Let $\mu$ be a probability measure on $\mathbb R$ and set

$$c(K):=\int_{\mathbb R}(x-K)^+d\mu(x).$$

Assume that one has a sequence of probability measures $(\mu_n)_{n\ge 1}$ s.t.

$$\int_{\mathbb R}\left(x-\frac{i}{2^n}\right)^+d\mu_n(x)=c\left(\frac{i}{2^n}\right) \mbox{ for all } -n2^n\le i\le n2^n.$$

It is easy to see that the sequence $(\mu_n)_{n\ge 1}$ admits a weakly convergent subsequence. Without loss of generality, we may assume that it is weakly convergent. Let $\mu_0$ be its limit. My question is that can we show that $\mu_0=\mu$? I believe that the answer is no, but I can't find a counterexample. Thanks for the reply!