Generalisation of Strassen's (Kellerer's) Theorem Let $\mu$ and $\nu$ be two probability measures on $\mathbb R^d$ with finite first movements, i.e.
$$\int_{\mathbb R^d}|x|~\mu(dx),\quad \int_{\mathbb R^d}|x|~\nu(dx) \quad <\quad +\infty.$$
$\mu$ and $\nu$ are said to be increasing in convex order, if the following inequality holds for any convex function $f$ with linear growth:
$$\int_{\mathbb R^d}f(x)~\mu(dx)\quad \le \quad \int_{\mathbb R^d}f(x)~\nu(dx).$$
Then Strassen's Theorem states that $\mu$ and $\nu$ are said to be increasing in convex order, iff there exists a martingale $(M,N)$ s.t. $M\sim\mu$ and $N\sim\nu$.
My question is the following: let $d=1$ and $\varepsilon\ge 0$. What are the conditions on $\mu$ and $\nu$ to ensure the existence of a stochastic process $(M,N)$ s.t. $M\sim\mu$, $N\sim\nu$ and 
$$ M~-~\varepsilon\quad \le \quad E[N|M]\quad \le \quad M~+~\varepsilon.\quad\quad\quad\quad\quad\quad (\ast)$$
Thanks a lot for the reply!
 A: I encourage you to consult Strassen's paper (http://www.jstor.org/stable/2238148). The result you quote as "Strassen's theorem" is Theorem 8 therein and follows from the much more general Theorem 7, as does the extension you seek. I'll sketch an argument:
By Theorem 7, a coupling as you describe exists if and only if  $\int \varphi\,d\nu \le \sup \int \varphi(y)\gamma(dx,dy)$ for every continuous $\varphi$ of linear growth, where the supremum is over all probability measures $\gamma$ on $\mathbb{R}^2$ of the form $\gamma(dx,dy) = \mu(dx)K_x(dy)$, where the kernel $K$ satisfies $\left|\int y\,K_x(dy) - x\right| \le \epsilon$ for $\mu$-a.e. $x$. Using a standard measurable selection argument, this is further equivalent to the statement that $\int \varphi\,d\nu \le \int h_{\varphi}\,d\mu$ for every continuous $\varphi$ of linear growth, where $h_{\varphi}(x) := \sup_{\eta \in M(x)}\int\varphi\,d\eta$ and where $M(x) := \sup\left\{\eta : \left|\int y\,\eta(dy) - x\right| \le \epsilon\right\}$.
To simplify this further, we find a better expression for $h_{\varphi}$. Note first that $h_{\varphi}(x) \ge \int\varphi\,d\delta_z \ge \varphi(z)$ for $|z-x| \le \epsilon$, so $h_{\varphi}(x) \ge \sup_{|z-x| \le \epsilon}\varphi(z)$. On the other hand, letting $\varphi^c$ denote the concave envelope, for each $\eta \in M(x)$ Jensen's inequality yields
$\int\varphi\,d\eta \le \int\varphi^c\,d\eta \le \varphi^c\left(\int y\,\eta(dy)\right) \le \sup_{|z-x| \le \epsilon}\varphi^c(z)$.
Thus, $h_{\varphi}(x) \le \sup_{|z-x| \le \epsilon}\varphi^c(z)$. It is a simple exercise to show (1) that $h_{\varphi}$ is concave and (2) that $x \mapsto \sup_{|z-x| \le \epsilon}\varphi^c(z)$ is concave and thus equal to the concave envelope of $x \mapsto \sup_{|z-x| \le \epsilon}\varphi(z)$. Hence, $h_{\varphi}(x) = \sup_{|z-x| \le \epsilon}\varphi^c(z)$.
Finally, we find that the your coupling exists if and only if 
$\int\varphi(x)\,\nu(dx) \le \int \sup_{|z-x| \le \epsilon}\varphi(z)\,\mu(dx)$
for every concave function $\varphi$ of linear growth.
