Suppose $L_1([0,1],\lambda)=L_1(\lambda)$ is the set of all $1$-integrable functions on $[0,1]$.
$$S=\{(f_1,f_2)\in L_1^2(\lambda) |0\leq f_1+f_2\leq 1, a.e. \}$$
By Dunford Pettis theorem, we know that for $j=1,2$,
$$S_j=\{f_j\in L_1(\lambda) |0\leq f_j\leq 1, a.e. \} $$
is weakly compact with respect to weak topology $\sigma(L_1(\lambda),L_{\infty}(\lambda))$, can I conclude that $S$ is also weakly compact with respect to weak topology $\sigma(L_1^2(\lambda),L_{\infty}^2(\lambda)$)?
No this set is not weak compact: The pair of constant functions $p_n=(n,-n)$ is in $S$, but certainly $(p_n)$ does not have a weakly converging subsequence (and the set cannot by weakly compact).