Question about the weak convergence of probability

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!

Let $c_n(k)$ be as $c(k)$ except for the measure $\mu_n$, $E_n$ be expectation under $\mu_n$ etc. . Since $| (x-a)^+ = (x-b)^+ | < |(a-b)|$, $c_n(k)$ is continuous. It is clearly also monotone. So it is easy to show that $c_n \rightarrow c$, even uniformly. Let $f$ be twice differentiable with compact support. $f(x) = \int f^{\prime \prime} (k) (x-k)^+ dk$ (by integrating by parts twice, so $E_n(f) = \int f^{\prime \prime} (k) c_n(k) dk$. It follows that $E_n(f) \rightarrow E(f)$ for such $f$. By choosing such a function that is 1 on $(-N,N)$ where $\mu(-N,N) \approx. 1$ it follows that most measures have most of their support in that interval. As any bounded continuous function can be uniformly approximated by a twice differentiable function with compact support over $(-N,N)$ it follows that the measures converge weakly.