Let $(\Omega,f,p)$ be a probability space. Let $L^{\infty}$ be the space of essentially bounded random variables and let $L^1 $ be the space of random variables $X$ with $E[|X|] < \infty.$ Since $(L^1,L^{\infty})$ forms a dual pair, $L^1$ can be considered with $\sigma(L^1,L^{\infty})$ the weak*-topology. $L^1$ also has the norm topology induced by the norm $\left\lVert X\right\rVert_1 =E[|X|] .$
If a set is closed with respect to the weak*-topology on $L^1,$ can we say that the set is closed with respect to the topology induced the norm $\left\lVert \boldsymbol{\cdot} \right\rVert_1 ?$
Any kind of answer, and related references are appreciated.
You're talking about the weak topology. (The terminology "dual pair" is unfortunate here, since $L^\infty=(L^1)^*$ but in general $L^1\ne(L^\infty)^*$.)
In any Banach space weakly closed implies norm-closed. Taking complements, this is the same as saying weakly open implies norm-open, which is clear from the definition. (This is why it's called the "weak" topology: It's weaker than the norm topology.)
(Or: If $E$ is weakly closed, $x_n\in E$ and $x_n\to x$ in norm then also $x_n\to x$ weakly, hence $x\in E$.)