Let
- $(X,d)$ be a complete separable metric space;
- $\mu$ be a Borel probability measure on $X$;
- $(f_n),(g_n)\subset L_1(X,\mu)$ be sequences of non-negative uniformly bounded sequences, bounded by $1$, i.e. for all $x\in X$ and for all $n$ we have
$$0\leq f_n(x),g_n(x)\leq 1,$$ $$||f_n||_1, ||g_n||_1\leq 1.$$
Suppose that $$f_n \rightarrow_{w^*} f\in L_1(X,\mu)$$ $$g_n \rightarrow_{w^*} g\in L_1(X,\mu)$$ $$f_n\times g_n\rightarrow_{w^*} F\in L_1(X,\mu)$$
Now the question is: is it true that $$F=f\times g?$$
(Throughout, $\times$ denotes pointwise multiplication.)
Discussion: If, under these assumptions, one can show that one can extract a subsequence that is almost surely convergent, then the result follows.
No. Take the underlying probability space to be $[0,1]$ with the uniform distribution. For each $n$, divide the unit interval into $2^n$ subintervals of equal length and number them consecutively. Let $f_n$ be the indicator function of the odd-numbered intervals and $g_n$ the indicator function of the even-numbered intervals. Then $f_n\times g_n=0$ for all $n$ and so is $F$. But the sequences $\langle f_n\rangle$ and $\langle g_n\rangle$ have the same weak-$\star$ limit, the function with constant value $1/2$ and $1/2\times 1/2=1/4\neq 0$ .