Martingale with respect to a sequence of real numbers. 
Let $\{c_n\}_{n=0}^\infty$ be a sequence of real numbers and let $\epsilon>0$. Then there exists a martingale $\{X_n\}_{n=0}^\infty$ such that 
  $$
\mathrm{P}(X_n = c_n\ \text{for all}\ n\geq 0)\geq 1-\epsilon.
$$

My attempt: it suffices to show that 
$$
\mathrm{P}(X_n \neq c_n\ \text{for some}\ n\geq 0)=\mathrm{P}\bigl(\bigcup_{n=0}^\infty \{ X_n\neq c_n\}\bigr)\leq \epsilon.
$$
So I want to construct a martingale $\{X_n\}_{n=0}^\infty$ such that for all $n$
$$
\mathrm{P}(X_n\neq c_n)\leq \frac{\epsilon}{2^{n+1}}.
$$
Any hint is appreciated.

Let $(\xi_j)_{j \in \mathbb{N}_0}$ be a sequence of independent random variables such that $$\mathbb{P}(\xi_j=m_j) = \frac{\epsilon}{2^{j+1}} \quad \text{and} \quad \mathbb{P}(\xi_j=c_j-c_{j-1}) = 1- \frac{\epsilon}{2^{j+1}}$$
where $m_j \in \mathbb{R}$ is chosen such that $$\mathbb{E}(\xi_j) = m_j \frac{\epsilon}{2^{j+1}} + (c_j-c_{j-1}) \left(1- \frac{\epsilon}{2^{j+1}} \right)=0.$$ (For $j=0$ put $c_{-1} :=0$) Show that $$X_n := \sum_{j=1}^n \xi_j$$ is a martingale. Deduce from $$\mathbb{P}(\forall j: \xi_j = c_j-c_{j-1}) \geq 1-\epsilon$$ that $$\mathbb{P}(\forall n: X_n = c_n) \geq 1-\epsilon.$$
