# Weak convergence preserver pointwise inequality

The proof of boundedness of Hardy-Littlewood maximal function in Sobolev spaces in Kinnunen's paper has the following argument:

"... Hence $(v_k)$ is a bounded sequence in $W^{1,p}(\mathbb{R}^n)$ which converges to $\mathcal{M}u$ pointwise. The weak compactness of Sobolev spaces implies $\mathcal{M} u \in W^{1,p}(\mathbb{R}^n), v_k$ converges to $\mathcal{M}u$ weakly in $L^p(\mathbb{R}^n)$ and $D_i v_k$ converges to $D_i \mathcal{M}u$ weakly in $L^p(\mathbb{R}^n)$. Since $|D_i v_k| \leq \mathcal{M} D_i u$ almost everywhere by (2.1), the weak convergence implies $|D_i \mathcal{M} u| \leq \mathcal{M} D_i u$ almost everywhere in $\mathbb{R}^n$"

I'm confused about the part where weak convergence implies pointwise boundedness. I know that weak convergence in $L^p$ preserves norm inequality but how does one get pointwise inequality?

Let $v \in L^2(\Omega)$ be arbitrary. The set $$\{ u \in L^2(\Omega) : u(x) \le v(x) \text{ f.a.a. } x \in \Omega\}$$ is (strongly) closed and convex. Thus, it is also weakly closed.