Weak limit of non-negative functions is non-negative (without Mazur)

Question

Let $\Omega \subseteq \mathbb R^2$ be compact subset.

Suppose that $g_n \ge 0$ lie in $L^1(\Omega)$ and that $g_n$ converges weakly in $L^1$ to $g$.

Is there a way to prove that $g \ge 0$ a.e. on $\Omega$ without using Mazur's lemma?

I guess what I have in mind is the following:

We have

$$\int_{\Omega} g f =\lim_{n \to \infty} \int_{\Omega} g_n f\ge 0 $$

for any $f \ge 0$ be in $L^{\infty}(\Omega)$.

Does this property imply that $g$ is non-negative? I think that there should be a way of showing this but I am not sure how...

Answer 1

Take $f = \mathbf{1}_{g\leq 0} \in L^\infty$. Then $$0\leq\int_\Omega gf = \int_{g \leq 0} g \leq 0.$$ So the function $gf$ is nonpositive and its integral is zero hence it vanishes a.e. This implies that $g \geq 0$ a.e.

Answer 2

Hint: Given any measurable $E\subset \Omega$ we have

$$\int_E g_n = \int_\Omega g_n\chi_E \to \int_\Omega g\chi_E = \int_E g.$$

Answer 3

Just chose $A=h^{-1}( \mathbb {R}_+) $ where h is a function in the equivalent class $g $. Then, the function $f $ defined to be the conjugate of $h $ divided by $|h|$ if $x \in A^c $ and $0$ otherwise gives you that $h $ is 0 almost evrywhere on $A^c $, so $h $ is in $\mathbb {R}_+$ almost everywhere.