Let $\{f_n\}, \{g_n\}$ be sequences of functions in $L^1_{[0,1]}([0,1])$ and $\lambda$ the Lebesgue measure on $[0,1].$ If $\{ \lambda(f_n^{-1}(A) \cap g_n^{-1}(B)) \}$ is a convergent sequence for all measurable sets $A, B,$ does there exists $f, g \in L^1_{[0,1]}([0,1])$ such that $\lambda(f_n^{-1}(A) \cap g_n^{-1}(B)) \to \lambda(f^{-1}(A) \cap g^{-1}(B))$ for all $A, B$?
My attempt: for the special case $B = [0,1],$ we want $\lambda(f_n^{-1}(A)) \to \lambda(f^{-1}(A))$ for all $A.$ Letting $h_1(A) = \lim\limits_{n \to \infty} \lambda(f_n^{-1}(A)),$ we can check $h_1$ is a measure. We want to find $f$ such that $\lambda \circ f^{-1} = h_1.$ It turns out there's infinitely many such $f,$ and once we find one then any other can be found from composing $f$ with a measure preserving (with respect to $\lambda$) bijection $[0,1] \to [0,1].$ Similarly defining $h_2(A) = \lim\limits_{n \to \infty} \lambda(g_n^{-1}(A)),$ there's infinitely many $g$ with $\lambda \circ g^{-1} = h_2.$
Thus, the problem reduces to the following: if $\lambda(f_n^{-1}(A)) \to \lambda(f^{-1}(A))$ and $\lambda(g_n^{-1}(A)) \to \lambda(g^{-1}(A))$ for all measurable sets $A,$ prove there exists measure preserving bijections $\mu_1, \mu_2 : [0,1] \to [0,1]$ such that $$\lambda(f_n^{-1}(A) \cap g_n^{-1}(B)) \to \lambda(\mu_1(f^{-1}(A)) \cap \mu_2(g^{-1}(B)))$$ for all $A, B.$
We have to satisfy an equation for all possible choices of 2 variables $(A, B)$ and have 2 degrees of freedom $(\mu_1, \mu_2)$ since $f, g$ are fixed, so I believe it's possible. However, it turns out that we have 1 redunancy since if $\nu = \mu_1^{-1} \circ \mu_2,$ then $$\lambda(\mu_1(f^{-1}(A)) \cap \mu_2(g^{-1}(B))) = \lambda(\mu_1^{-1}(\mu_1(f^{-1}(A)) \cap \mu_2(g^{-1}(B)))) = \lambda(f^{-1}(A) \cap \nu(g^{-1}(B))).$$ Recall that in finding $f$ or $g$ individually, we had the same redundancy since $\lambda \circ f^{-1} = \lambda \circ \mu \circ f^{-1}$ for any measure preserving map $\mu.$ My issue at this point is that although I can choose $\nu$ for any particular choice of $(A, B),$ I'm not sure how to choose a $\nu$ which works for all $A, B.$ Any ideas?
$\def\CC{\mathcal{C}}\def\BB{\mathcal{B}}\def\N{\mathbb{N}}\def\R{\mathbb{R}}\def\menos{\smallsetminus}\def\vacio{\varnothing}\let\leq=\leqslant\let\geq=\geqslant\def\lim{\mathop{\mathrm{lim}}\limits}$Let $h_n := (f_n, g_n)$. We define the pre-measure on the semiring $\{A \times B : A, B \in \BB\}$, where $\BB$ is the set of Borel subsets of $[0,1]$, by $$\mu(A \times B) := \lim_n \lambda(h_n^{-1}(A \times B)).$$ We have to verify that $\mu(\vacio) = 0$, clear, and that $\mu(A \times B) = \sum_{k\in\N} \mu(A_k \times B_k)$ if $A \times B = \bigcup_{k\in\N} A_k \times B_k$ is a disjoint union. This follows easily from the dominated convergence theorem. So by Carathéodory's extension theorem $\mu$ extends to a Borel probability measure on $[0,1]^2$.
We want $h : [0,1] \to [0,1]^2$ Borel measurable such that $\lambda(h_n^{-1}(A \times B)) \to \lambda(h^{-1}(A \times B))$ for every $A, B \in \BB$ and $h = (f, g)$ with $f, g \in L^1$. It's generally true that if $\mu$ is a Borel probability measure on a Borel space $X$ then there is $\phi : [0,1] \to X$ such that $\lambda(\phi^{-1}(E)) = \mu(E)$ for every $E \subset X$ Borel, and $[0,1]^2$ is a Borel space (see Kallenberg, Foundations of Modern Probability, Lemma 3.2 (vii)). So $h$ such that $\lambda(h_n^{-1}(A \times B)) \to \lambda(h^{-1}(A \times B))$ for every $A, B \in \BB$ exists. It's easy to see that this extends to $A,B$ measurable.
We have show that if $h=(f,g)$ then $f,g \in L^1$, but $\lambda(h^{-1}([0,1]^2)) = \lim_n \lambda(h_n^{-1}([0,1]^2)) = 1$, so $f, g : [0,1] \to [0,1]$ and therefore they are in $L^1$.
I hope I'm understanding the hypotheses correctly.