TITLE: Maximum of measures over sets and functions
QUESTION [8 upvotes]: Let $(X,\mathcal A)$ be any measurable space and denote by $\mathrm b\mathcal A_1$ the set of all real-valued $\mathcal A$-measurable functions $f$ satisfying $\|f\|:=\sup_{x\in X}|f(x)|\leq 1$.
Let $P$ and $Q$ be two arbitrary families of probability measures on $(X,\mathcal A)$ and denote 
$$
  P^*f := \sup_{p\in P} \int_X f\,\mathrm dp
$$ 
for any $f\in\mathrm b\mathcal A_1$, and similarly denote $Q^*$. Let for simplicity write $P^*(A) = P^*1_A$ for any $A\in \mathcal A$ where $1_A:X\to \{0,1\}$ is the indicator function of the set $A$.  Is that true that
$$
  Q^*f\leq P^*f \quad \forall f\in \mathrm b\mathcal A_1 \iff Q^*(A)\leq P^*(A) \quad \forall A\in \mathcal A.
$$
The direction $\implies$ is obvious, however I'm not sure what about the other direction. I'd be also happy if you can suggest the source where similar properties of $P^*$ are considered. For what I know now, it shall be a capacity, but that's pretty much it.

REPLY [2 votes]: Take $X = \{1,2,3\}$,  $P = \{\delta_1, \frac12\delta_2+\frac12\delta_3\}$, $Q=\{\frac12\delta_1+\frac12\delta_2\}$ and $f$ defined as $f(1) = 0$, $f(2)=1$ and $f(3)=-1$. Then $$Q^*(A)\leq P^*(A) \quad \forall A\in \mathcal A,$$
but
$$P^*f = 0 < \frac12 = Q^*f.$$