Brezis' Exercise 4.10.3: can we pick such a measurable $u_n$?

Question

I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,

Let $(\Omega, \mathcal F, \mu)$ be a measure space with $\mu(\Omega) < \infty$. Let $p \in [1, \infty)$ and $j:\mathbb R \to \mathbb R$ be convex. We define $$ J:L^p(\Omega) \to (-\infty,+\infty], u \mapsto \int_\Omega j \circ u. $$ Let $p'$ be the Hölder conjugate of $p$, i.e., $\frac{1}{p} + \frac{1}{p'} =1$. Let $j^*:\mathbb R \to (-\infty,+\infty]$ and $J^*: L^{p'}(\Omega) \to (-\infty,+\infty]$ the convex conjugates of $j$ and $J$ respectively. Prove that $$ J^*(f) = \int j^* \circ f \quad \forall f \in L^{p'}(\Omega). $$

My below attempt would be complete if I could pick such a measurable $u_n$. Could you elaborate on how to finish my proof?

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My attempt Notice that $j$ is continuous, so $j^*$ is measurable. Fix $f \in L^{p'}(\Omega)$. By definition, $$ J^*(f) := \sup_{u \in L^p (\Omega)} \int ( fu - j \circ u). $$

Clearly, $$ f(\omega) u(\omega) - j (u (\omega)) \le \sup_{t \in \mathbb R} (f(\omega) t- j(t)) = j^* \circ f(\omega) \quad \forall \omega \in \Omega. $$

Hence $J^*(f) \le \int j^* \circ f$. Let's prove the reverse. WLOG, we assume $J^*(f)$ is finite. Let $B_n := \{t \in \mathbb R : |t| \le n\}$. Then $B$ is compact. Notice that $j$ is continuous. We define $$ \varphi_n:\mathbb R \to (-\infty, +\infty], x \mapsto \max_{t \in B_n} (x t- j(t)). $$

Then for every $x \in \mathbb R$, we have $\varphi_n (x) \uparrow j^* (x)$ as $n \to \infty$. For each $(n, \omega) \in \mathbb N^* \times \Omega$, there is (not necessarily unique) $u_n (\omega) \in B_n$ such that $$ \varphi_n \circ f(\omega) = f(\omega) u_n (\omega) - j(u_n (\omega)). $$

Clearly, $u_n:\Omega \to \mathbb R$ is bounded. We now assume $u_n$ is measurable. Because $\mu(\Omega) < \infty$, we get $u_n \in L^p (\Omega)$. So $$ \int \varphi_n \circ f = \int ( fu_n - j \circ u_n) \le J^* (f). $$

We have $\varphi_n \circ f \in L^1 (\Omega)$ and $\varphi_n \circ f \le \varphi_{n+1} \circ f$ for all $n$. By monotone convergence theorem, $$ \int j^* \circ f = \lim_n \int \varphi_n \circ f \le J^* (f). $$

This completes the proof.

Answer 1

It turns out that such measurable selection is possible in our case. It follows from below theorem, i.e.,

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Let $X$ be a polish space and $(S,\Sigma)$ a measurable space. Let $\varphi: S \twoheadrightarrow X$ be a weakly measurable correspondence with nonempty compact values, and suppose that $f: S \times X \to \mathbb{R}$ is a Caratheodory function (measurable in $s$ and continuous in $x$). Define the value function $m$ by: $$ m(s) = \max_{x \in \varphi(s)} f(s,x), $$ and the correspondence $$ \mu(s) = \{ x \in \varphi(s): f(s,x) = m(s) \}. $$ Then:

• The value function $m$ is measurable.
• The argmax correspondence $\mu$ is measurable, has nonempty compact values and admits a measurable selector.

Clarification: By weakly measurable correspondence is meant that $\{s \in S: \varphi(s) \cap K \neq \emptyset \}$ belongs to $\Sigma$ for each compact set $K$.