Follow up question to this one.
It is later proved that defining $\phi : C(Q)^* \to X$ as
$$
\phi(\mu) = \int_Q x d \mu (x)
$$
if $P$ is the set of all regular probability measures on $Q$ we have $ H \subset \phi(P) \subset \overline{H}$, therefore to prove the equality we have to show that $\phi(P)$ is closed in $X$. One fact needed to show this is
(i) P is weak*-compact in $C(Q)^*$
The proof starts by noticing
$$
P \subset \left\{ \mu : \left| \int_Q h d\mu \right| \leq 1 \; \text{if} \; \lVert h \rVert < 1 \right\}
$$
As usual I'm sure the observation is very trivial and I'm missing it…
Could you please elaborate the details explaining the inclusion?
All the rest concerning (i) seems fine.
Update:
I think if we assume $\lVert h \rVert < 1$ we can write
$$
1 = \mu(Q) > \lVert h \rVert < 1 \mu(Q) = \int_Q \lVert h \rVert d \mu > \int_Q \left| h \right| d \mu > \left| \int_Q h d \mu \right|
$$
Bearing in mind that every integral involved in this chain of inequalities is a standard lebesgue integral.
Best Answer
Assuming that $\|h\|$ denotes the $\sup$-norm of $h$, we have that $$\bigg |\int_Q h d\mu \bigg| \leq \int_Q |h| d \mu \leq \int_Q \|h\| d\mu = \|h\| \mu(Q) = \|h\|$$ for $\mu \in P$, since $P$ contains only probability measures on $Q$. Hence, if $\|h\| < 1$ and $\mu \in P$ then $$\bigg |\int_Q h d\mu \bigg| \leq 1$$ as desired.