For two probability measure $\pi_1=\frac{\delta_0+\delta_1}{2}$ and $\nu=\frac{\delta_0+\delta_1+\delta_2}{3}$, why do we have the 1-Wasserstein distance $$ W_1(\pi_1,\nu)=\frac{1}{2}? $$
Also, for $\pi_2=\delta_2$, why do we have $W_1(\pi_2,\nu)=1$? What is the result of the following integral for $x_1=1,2$, $$ \int W(\pi_{x_1},\nu)d\mu(x_1) $$ Is it equal to $1\times W(\pi_1,\nu)+2\times W(\pi_2,\nu)=\frac{1}{2}+2?$
First, consider the simpler case $W_1(\pi_2, \nu) = 1$. Consider the coupling $(X, Y)$, where we take $X \sim \pi_2$ and $Y \sim \nu$ independent of each other. Then because $X = 2$ almost surely, we have $\mathbb{E}|X - Y| = 1 \times \nu(1) + 2 \times \nu(0) = \frac{1}{3} + \frac{2}{3} = 1$. Since $W_1(\pi_2, \nu)$ is the infimum over all such couplings $(X, Y)$ of $\mu$ and $\nu$, we have $W_1(\pi_2, \nu) \leq 1$. To obtain a lower bound of the same value, consider moving all the mass of $\pi_2$ (which is concentrated on $2$) to obtain $\nu$ (which is uniform over $\{0, 1, 2\}$). Intuitively, we must move at least $\frac{2}{3}$ mass to $\{0, 1\}$, and since $1$ has capacity $\frac{1}{3}$, we must move at least $\frac{1}{3}$ mass to $0$. So any coupling $(X, Y)$ must have $\mathbb{E}|X - Y| \geq 1$. Thus $W_1(\pi_2, \nu) = \inf_{(X, Y) \sim (\pi_2, \nu)} \mathbb{E}|X - Y| = 1$.
A similar argument yields $W_1(\pi_1, \nu) = \frac{1}{2}$. We can exhibit the optimal coupling (which gives the upper bound). Let $\mathbb{P}(X = Y = 0) = \frac{1}{3}$ and $\mathbb{P}(X = Y = 1) = \frac{1}{3}$; i.e. the minimum over the common support of $\pi_1$ and $\nu$. To fill in the remaining mass, let $\mathbb{P}(X = 0, Y = 2) = \mathbb{P}(X = 1, Y =2) = \frac{1}{6}$. Under this coupling, we have $\mathbb{E}|X - Y| = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} = \frac{1}{2}$. An analogous argument as the simpler case above shows that this is also the matching lower bound. Hence $W_1(\pi_1, \nu) = \frac{1}{2}$.
The result of your integral depends on what $\mu$ is (apparently some measure supported on $\{ 1, 2 \}$): $\int W(\pi_{x_1}, \nu) \,\mathrm{d} \mu(x_1) = W(\pi_1, \nu) \cdot \mu(1) + W(\pi_2, \nu) \cdot \mu(2)$.