Let $(S,\Sigma)$ be a measurable space and $\mu$ an arbitrary probability measure for $(S,\Sigma)$. Consider the set of all probability measures for that space with same mean $p$:
$\mathcal{S}(p)=\{\mu\in(S,\Sigma): \int_{S} s\mu(ds)=p\}$.
Endow this measurable space with the first-order stochastic dominance order, that is, $\mu'\succsim_{FOSD}\mu$ if for every increasing, bounded function $f: S\to\mathbb{R}$, it holds that $\int_{S}f(s)\mu'(ds)\ge \int_{S}f(s)\mu(ds)$.
Question: is $\mathcal{S}(p)$ a lattice?
$\mathcal S _(p)$ is not a lattice unless mistaken.
Take $\mu$ a measure loading points $0$ and $1$ with probability $1/2$ and $\mu'$ loading points $1/4$ and $5/8$ respectively with probabilities $1/3$ and $2/3$. Then $\mu \wedge \mu'$ loads point $0 $ with mass $1/2$ then point $1/4$ at another $1/3$ probability and finally point $1$ with what's left so at $1/6=(1-(1/2+1/3))$ Now let's integrate $\mu$ and $\mu'$ and $\mu \wedge \mu'$ over $X$ $$E_\mu[X]=1/2$$ $$E_\mu'[X]=1/3*1/4 +2/3*5/4 =1/2$$ So they both are in $\mathcal S_(1/2)$
But $E_{\mu'\wedge \mu} [X]=1/3*1/4+ 1/6*1=1/4$ is in $\mathcal S _(1/4)$
So it's not a lattice I took the definition of $\mu \wedge \mu'(s)=max(\mu(s),\mu'(s))$