Let $g:(0,1)\rightarrow \Bbb{R}$ and define $$Q_g:=\left\{f\in L^1((0,1)): |f(x)|\leq |g(x)|~~\text{for almost every}~~x\in (0,1)\right\}\subset L^1\left((0,1)\right)$$ I want to show that this set is weakly closed in $L^1((0,1))$.
My idea was to use sequences. So I wanted to pick $(f_n)_n\subset Q_g$ a sequence s.t. $f_n\stackrel{\text{weakly}}{\longrightarrow} f\in L^1((0,1))$. I need to show that $|f(x)|\leq |g(x)|$ for almost every $x$ to conclude that $Q_g$ is closed. Since $f_n\stackrel{\text{weakly}}{\longrightarrow} f$ I know that for all $L\in \left(L^1((0,1))\right)^*$ we have $L(f_n)\rightarrow L(f)\Leftrightarrow |L(f_n-f)|\rightarrow 0$. Now I wanted to look at $$\begin{align}\big||f(x)|-|g(x)|\big|&\leq |f(x)-g(x)|\\&\leq |f(x)-f_n(x)|+|f_n(x)-g(x)|\end{align}$$But now the problem is that I don't know if $|f_n(x)-g(x)|\leq 0$ I only know that $|f_n(x)|-|g(x)|\leq 0$. And I don't see how to fix it. Could maybe someone give me a hint?
Continuing from my comment, you can reduce yourself to show that $\{f\in L^1\; :\;f(x)\geq 0\}$ is weakly closed. So, take $(f_n)_n$ a sequence of positive functions, $f_n\to f$ weakly. For any $h\in L^\infty$, $h\geq 0$, $0\leq L_h(f_n)\to L(f)$ as $n\to \infty$ because $f_n$ are positive, where $L_h(\phi):=\int h\phi\;dx$.
So, $\int h\phi\;dx\geq 0$ for any $h\in L^\infty$. This necessarily implies $f\geq 0$.