Given a statistical model $(X,\mathcal{F},(\mathbb{P}_\vartheta)_{\vartheta\in\Theta}$) and $\hat{\vartheta}$ be an estimator. Then (here the MSE of $\hat{\vartheta}$ in terms of $\vartheta$ is denoted as $R(\hat{\vartheta},\vartheta)$.
$$ \sup_{\vartheta\in \Theta}R(\hat{\vartheta},\vartheta)= \sup_\Pi \int_\Theta R(\hat{\vartheta},\vartheta)\Pi (d\vartheta) $$ Where the supremum is taken over all a-priori distributions $\Pi$. Now what is clear to me ist that $$ R(\hat{\vartheta},\vartheta)\leq \sup_\Pi \int_\Theta R(\hat{\vartheta},\vartheta)\Pi (d\vartheta) $$ This can be shown using the Dirac-measure as one of the a-priori measures in $\Pi$. What is not clear to me is the other way, where the proof only says $$ \text{It is clear that } \sup_{\vartheta\in \Theta}R(\hat{\vartheta},\vartheta)\geq \int_\Theta R(\hat{\vartheta},\vartheta)\Pi (d\vartheta) $$ How can this be shown?
I think applying Hölder's inequality proves the claim. ($p=\infty$ corresponds to supremum, and $q=1$ results in 1 integrating over the prior.) But I cannot see how "it is clear" either!
EDIT
On a second thought, it is clear. Suppose $f(x) < M$. Then $\int f(x)g(x)dx < M \int g(x)dx $. Here, $M$ is the supremum of MSE, and $ \int \Pi(d\vartheta) = 1 $.