$ \int |y-x| d \mu (y) \leq \int |y-x| d \nu (y)$ then for $\phi$ convex $ \int \phi(y) d \mu (y) \leq \int \phi(y) d \nu (y)$ How to prove that for two measures with finite moments,
$$ \forall x \in \mathbb R,  \quad \int |y-x| d \mu (y) \leq  \int |y-x| d \nu (y)$$
$ \qquad \qquad \implies \forall \phi, \text{ convex function, we have that }$
$$\int \phi(y) d \mu (y) \leq  \int \phi(y) d \nu (y) $$
Perhaps there should also be the condition: $$ \int d \mu = \int d \nu $$
but for this I am not sure. That is the reason why I would like to find the proof.

My attempt.
I tried to approximate any convex function with functions like $(y-x)^+$ but it is hard to get jumps with continuous function like this one. I was thinking of multiplying by a constant $(y-x)^+$  to get a steeper slope, and combining $(y-x_0)^+$  with $(y-x_1)^+$  in order to get a constant function over the interval $[x_0, x_1]$, but it seems to me it is not exactly the right approach.

Source.
It is from the book Peacocks and Associated Martingales, with Explicit Constructions, exercise 1.7. The topics related are ordering of measures, convex order of measures...
 A: After multiple thoughts, I think I have settle down this problem. Here is my approach:
We work under the measure $\mu$.
First we prove that we can  approximate any characteristic function by a sequence of linear combination of Call function.
We prove the result for real valued functions, since transferring the problem to multi-dimensional spaces is straight-forward. We only prove the statement for sets of the form : $[a, \infty[$ and $a \in \mathbb  R$.  This proves for any set of the Borel set of $\mathbb R$ using the $\pi-\lambda$ theorem.
\vspace{0.2cm}
The proof goes as follows: first we fix $a, c \in \mathbb R$ and we are searching for a sequence of function $\phi_n$, constructed with Call functions such that $\phi_n$ approximates $c \cdot 1_{ [a,\infty[}$.
We set $\forall n\in \mathbb N, \phi_n(x) = c \cdot n \left [ ( x - (a - \frac 1 n))^+ - (x - a)^+ \right ]$.
Then, the error of the approximation is given by:
\begin{align*}
\int_{\mathbb R^d} | c \cdot 1{ [a,\infty[} - \phi_n | d \mu &= 
\int_{a - \frac 1 n}^{a} | c - c \cdot n  [  x - a + \frac 1 n ] | d \mu \\
&= \int_{a - \frac 1 n}^{a} | c \cdot n ( x - a ) | d \mu \\
&=  \frac c {2n} .
\end{align*}
Then, $\phi_n$ converges to $1_{ [a,\infty[}$ in $L^1 (d  \mu)$. Now, since any positive measurable function can be approximated by a sequence of linear combination of simple functions, we can approximate any positive measurable function by a linear combination of Call functions.
Then, it is easy to prove the statement with dominated convergence :)
