is closed in the original topology [@FoellSch; @2004 Thm A.59].

\[static minimal penalty funct. in Q(&lt;&lt;) &lt;=&gt;\] Let $\psi :\mathcal{K}\subset \mathcal{Q}_{\ll }(\mathbb{P})\rightarrow \mathbb{R}\cup \{+\infty
\} $ be a function with $\inf\nolimits_{\mathbb{Q}\in \mathcal{K}}\psi (\mathbb{Q})\in \mathbb{R},$ and define the extension $\psi (\mathbb{Q}):=\infty $ for each $\mathbb{Q}\in \mathcal{Q}_{cont}\setminus \mathcal{K}$, with $\mathcal{K}$ a convex closed set. Also, define the function $\Psi $, with domain in $L^{1}$, as $$\Psi \left( D\right) :=\left\{
\begin{array}{rl}
\psi \left( \mathbb{Q}\right) & \text{if }D=d\mathbb{Q}/d\mathbb{P}\text{
for }\mathbb{Q}\in \mathcal{K} \\
\infty & \text{otherwise.}\end{array}\right.$$Then, for the convex measure of risk $\rho (X):=\sup\limits_{\mathbb{Q}\in
\mathcal{Q}_{cont}}\left\{ \mathbb{E}_{\mathbb{Q}}\left[ -X\right] -\psi
\left( \mathbb{Q}\right) \right\} $ associated with $\psi $ the following assertions hold:

$\left( a\right) $ If $\rho $ has as minimal penalty $\psi _{\rho }^{\ast }$ the function $\psi $ (i.e. $\psi $ $=\psi _{\rho }^{\ast }$ ), then $\Psi $ is a proper convex function and lower semicontinuous w.r.t. the (strong) $L^{1}$-topology or equivalently w.r.t. the weak topology $\sigma \left(
L^{1},L^{\infty }\right) $. $\left( b\right) $ If $\Psi $ is lower semicontinuous w.r.t. the (strong) $L^{1}$-topology or equivalently w.r.t. the weak topology $\sigma \left(
L^{1},L^{\infty }\right) ,$ then $$\psi \mathbf{1}_{\mathcal{Q}_{\ll }(\mathbb{P})}=\psi _{\rho }^{\ast }\mathbf{1}_{\mathcal{Q}_{\ll }(\mathbb{P})}.
\label{PSI_l.s.c=>psi*=psi_on_Q<<}$$

*Proof:* $\left( a\right) $ Recall that $\sigma \left(
L^{1},L^{\infty }\right) 