Consider the space $M$ of Borel measures on the real unit interval $[0,1]$, equipped with the 1-Wasserstein metric $d_W$ (or "Earth mover's distance"). The expected value is then a map $M\to [0,1]$ given by: $$ p \mapsto E_p := \int_{[0,1]} x \, dp(x)\;. $$
Is it true that this map is Lipschitz? That is, can we find a constant $C$ such that for every $p,q\in M$, $|E_p-E_q|\le C\,d_W(p,q)$ ?
Let $p$ and $q$ be two arbritrary probability measures on $[0,1]$. By definition,
$$d_W(p,q) = \inf_{\gamma \in \Gamma_{p,q}} \int |x-y| \, d\gamma(x,y)$$
where $\Gamma_{p,q}$ is the set of Borel measures on $[0,1] \times [0,1]$ with marginals $p$ and $q$. As $p$ and $q$ are probability measures, we have for any $\gamma \in \Gamma_{p,q}$
$$\int_{[0,1]} x \, dp(x) = \int_{[0,1] \times [0,1]} x \, d\gamma(x,y) \quad \text{and} \quad \int_{[0,1]} y \, dq(y) = \int_{[0,1] \times [0,1]} y \, d\gamma(x,y).$$
Hence,
$$\begin{align*} |E_p-E_q| &= \left| \int_{[0,1] \times [0,1]} (x-y) \, d\gamma(x,y) \right|\leq \int_{[0,1] \times [0,1]} |x-y| \, d\gamma(x,y). \end{align*}$$
Since this inequality holds for any $\gamma \in \Gamma_{p,q}$ we can take the infimum over all $\gamma \in \Gamma_{p,q}$ to conclude
$$|E_p-E_q| \leq d_W(p,q).$$
For $(M,d)$ compact, let $I_f(\mu):=\int_M f\,d\mu$. The Kantorovich-Rubinstein formula states that,
$$ W_1(\mu,\nu) =\!\!\!\sup_{f\in\operatorname{Lip}_1(M,\mathbb R)} |I_f(\mu)-I_f(\nu)|, $$
Immediately we see that if $f$ is 1-Lipschitz then so is $I_f$ because $|I_f(\mu)-I_f(\nu)|\leq W_1(\mu,\nu)$. Moreover, if $f$ is $K$-Lipschitz then so is $I_f$ (simply substitute $g=\tfrac{1}{K}f\in\operatorname{Lip}_1(M,\mathbb R)$).
For your question, let $M=[0,1]$ and $f(x)=x$. Then $E_p=I_f(p)$ is 1-Lipschitz because $f$ is.
