Prove that $C$ is closed in the weak$^*$ topology $\sigma(L^\infty, L^1)$

Question

Let $(\Omega, \mathcal F, \mu)$ be a $\sigma$-finite measure space. I'm trying to solve below exercise, i.e.,

Let $f:\Omega \to \mathbb R$ be measurable and $p \in [1, \infty]$. Let $$ C := \{u \in L^p (\Omega) : u \ge f \text{ a.e.}\}. $$

1. Let $p < \infty$. Prove that $C$ is closed in the weak topology $\sigma(L^p, L^{p'})$ where $p'$ is the Hölder conjugate of $p$.
2. Let $p = \infty$. Prove that $$ C= \left \{u \in L^\infty (\Omega) : \begin{aligned} \int u \varphi \ge \int f \varphi \text{ for all } \varphi \in L^1(\Omega) \\ \text{s.t. } f \varphi \in L^1 (\Omega) \text{ and } \varphi \ge 0 \text{ a.e.} \end{aligned} \right \}. $$
3. Let $p = \infty$. Deduce that $C$ is closed in the weak$^*$ topology $\sigma(L^\infty, L^1)$.
4. Let $f_1, f_2 \in L^\infty (\Omega)$ with $f_1 \le f_2$ a.e. Prove that the set $$ E := \{u \in L^\infty (\Omega) : f_1 \le u \le f_2 \text{ a.e.}\} $$ is compact in $L^\infty (\Omega)$ w.r.t. the weak$^*$ topology $\sigma(L^\infty, L^1)$.

Could you confirm if my attempt on the sub-question (2.) is fine?

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1. It's clear that $C$ is convex. It suffices to prove that $C$ is closed in $L^p (\Omega)$. Let $(u_n) \subset C$ and $u \in L^p (\Omega)$ such that $u_n \to u$ in $L^p$. There is a subsequence $\varphi$ of $\mathbb N$ such that $u_{\varphi (n)} \xrightarrow{n \to \infty} u$ a.e. We have $u_{\varphi (n)} \ge f$ a.e. for all $n$. By taking the limit $n \to \infty$, we get $u \ge f$ a.e.

2. Let $$ D := \left \{u \in L^\infty (\Omega) : \begin{aligned} \int u \varphi \ge \int f \varphi \text{ for all } \varphi \in L^1(\Omega) \\ \text{s.t. } f \varphi \in L^1 (\Omega) \text{ and } \varphi \ge 0 \text{ a.e.} \end{aligned} \right \}. $$

It's clear that $C \subset D$. Let's prove the reverse, i.e., $D \subset C$. Assume the contrary that there is $\overline u \in D \setminus C$. First, $\mu (\{\overline u-f<0\}) >0$. By continuity of measure from below, there is $\overline n \in \mathbb N$ and $\varepsilon>0$ such that $\mu (\{-\overline n<\overline u-f< - 1/\overline n\}) >\varepsilon$. By $\sigma$-finiteness of $\mu$, there is $E \in \mathcal F$ such that $E \subset \{-\overline n<\overline u-f< - 1/\overline n\}$ and $0 <\mu(E) < \infty$. Second, let $\overline \varphi := 1_E$. Clearly, $\overline \varphi,f \overline \varphi \in L^1 (\Omega)$. We have $\int (\overline u - f) \overline\varphi \le - \mu(E) / \overline n<0$, which is a contradiction.

3. Let $(u_n) \subset D$ and $u \in L^\infty (\Omega)$ such that $u_n \to u$ in $\sigma(L^\infty, L^1)$. We fix $\varphi \in L^1(\Omega)$ such that $f \varphi \in L^1 (\Omega)$ and $\varphi \ge 0$ a.e. Then $\int u_n \varphi \to \int u \varphi$. On the other hand, $\int u_n \varphi \ge \int f \varphi$ for all $n$. It follows that $\int u \varphi \ge \int f \varphi$. Hence $u \in D$.

4. Clearly, $E$ is bounded in $\| \cdot \|_\infty$. By (3.), $E$ is closed in $\sigma(L^\infty, L^1)$. The claim then follows from Banach–Alaoglu theorem.