**Setup**
---------


Suppose that $U$ is a subset of $L^{\infty}\_{\mu}(\mathscr{F})\cap L^1\_{\mu}(\mathscr{F})$ defined by 
$$
f\in U \Leftrightarrow g(f(x))\leq M \mbox{ and } f \in L^{\infty}\_{\mu}(\mathscr{F})\cap L^1\_{\mu}(\mathscr{F})
,
$$
for a fixed convex and lower-semicontinuous function $g$
. Here $\mu$ is a (if acceptable $\sigma$-)finite measure on $\mathcal{X}$. 


**Question**
------------


Under what conditions can the interchanging of the 
$$
\inf\_{f \in U}\int\_{x \in \mathcal{X}}f^2(x)\mu(dx)
 = 
\int\_{x \in \mathcal{X}}\inf\_{f \in U}f^2(x)\mu(dx).
$$
In other words....when can the infimum be taken pointwise?



Echange of Infimum Integral with Pointwise Infimum
If $U$ is non-empty then the value of either side is $m \mu(\mathcal{X})$ where $m= \inf \{y^2 ~|~ g(y)<M\}$. If $\mu$ is non-finite then $m=0$. 


To see this note first that for $f\_1,f\_2 \in U$ the function $f$ defined by 
$$
f(x)=\begin{cases}
f\_{1}(x) & f\_{1}^{2}(x)\le f\_{2}^{2}(x)\\
f\_{2}(x) & f\_{2}^{2}(x)<f\_{1}^{2}(x).
\end{cases}
$$
is in $U$ as well. Also note $\int f^2 d\mu \le \int f\_i^2d\mu$.


A second observation: if $U$ is non-empty then then either $f\_{min} \equiv \sqrt{m}$ or $f\_{min} \equiv -\sqrt{m}$ is in $U$ as $m \le f^2$ for all $f$ and either $g(\sqrt{m}) \le M$ or $g(-\sqrt{m}) \le M$ by definition of $m$. Note also $f\_{min}^2(x) = \inf\_{f\in U} f^2(x)$.


Combining those two facts one gets 
$$\inf\_{f\in U} \int f^2 d\mu = \int \inf\_{f\in U} f^2 d\mu = \int f\_{min}^2 d\mu = m \mu(\mathcal{X}).$$ 


Side remark: This does not need any assumptions on $g$ except beeing somewhere finite-valued and $g(0) \le M$ for non-finite measures.


