Prove an inequality using Jensen's inequality

Question

Let $Q\in M_1(M_1(\Sigma)),$ whereby $M_1(\Sigma)$ denotes the space of probability measures on $\Sigma.$ We define a new measure $\mu_Q (\Gamma) = \int_{M_1(\Sigma)} \nu(\Gamma)Q(d\nu), \Gamma \in B_{\Sigma},$ the Borel sigma algebra on $\Sigma, \nu \in M_1(\Sigma).$ Let $V: \Sigma \rightarrow [0, \infty]$ be bounded and measurable function. I need to prove the following inequality $$\int_{M_1(\Sigma)} \exp[\int_{\Sigma} Vd\nu] Q(d\nu)\ \leq \int_{\Sigma} \exp[V] d\mu_Q.$$ Given that the exponential is convex, I use the Jensen's inequality to get for the inner integral, $$\exp[\int_{\Sigma} V d\nu] \leq \int_{\Sigma} \exp[V] d\nu.$$ Inserting this into the integral expression we get, $$\int_{M_1(\Sigma)} \exp[\int_{\Sigma} Vd\nu] Q(d\nu)\ \leq\int_{M_1(\Sigma)}\int_{\Sigma}\exp[V] d\nu Q(d\nu).$$ I do not know how to go from here. I guess one may write $d\mu_Q = \nu Q(d\nu),$ given the integral definition of $\mu_Q.$ But inserting $Q(d\nu) = \frac{1}{\nu} d\mu_Q,$ is not providing the result. One should very probably use Fubini, but even this did not convince me how to correctly provide the final result. Can somebody help ? Thanks.

Answer 1

I think that the problem is only superficial. By that I mean that you defined $\mu_Q := \int \nu Q(d\nu)$, so that heuristically you get exactly $$ \int fd\mu_Q = \int \int fd\nu\;Q(d\nu)\qquad\text{ for all }f\in L^1(\mu_Q). $$ The rest of this answer is concerned with making this rigorous.

The definition of your measure immediately implies that $$ \int_\Sigma 1_A d\mu_Q = \mu_Q(A) = \int \nu(A)\;Q(d\nu) = \int \int_\Sigma 1_A(x) d\nu(x) \;Q(d\nu). $$ This means immediately that $$ \int_\Sigma fd\mu_Q = \int \int_\Sigma f(x)d\nu(x)\;Q(d\nu) $$ whenever $f$ is a simple function (i.e. $f= \sum_{k=1}^n a_k 1_{A_k}$ with $A_1,\dots, A_n$ disjoint).

From here, you continue as one did for the construction of the integral. First, you take a positive function $f$ and approximate it pointwisely (that means everywhere and not a.e.) by an increasing sequence of the previous simple functions $f_n$. Then, by using monotone convergence multiple times, one has \begin{align*} \int_\Sigma f d\mu_Q = \lim_n \int_\Sigma f_n d\mu_Q &= \lim_n \int \int_\Sigma f_n(x) d\nu(x)\; Q(d\nu) \\ &= \int \lim_n \int_\Sigma f_n(x)d\nu(x)\;Q(d\nu) \\ &= \int \int_\Sigma f(x)d\nu(x)\;Q(d\nu). \end{align*} For general $f\in L^1(\mu_Q)$, you consider the positive and the negative real part separately to conclude that $$ \int fd\mu_Q = \int \int fd\nu \;Q(d\nu). $$