Let $\{f_n\}$ be a sequence of functions in $L^1_{[0,1]}([0,1])$ and $\lambda$ the Lebesgue measure on $[0,1].$ If $\{ \lambda(f_n^{-1}(U)) \}$ is a convergent sequence for all measurable sets $U,$ does $\{f_n\}$ have a convergent subsequence in $L^1$?
My attempt: let $f(U) = \lim\limits_{n \to \infty} \lambda(f_n^{-1}(U)),$ then it is easy to check $f$ is a measure on $[0,1].$ If $f_n \to g,$ we need $\mu(g^{-1}(U)) = f(U)$ for all $U.$ However, this equation only determines $g$ up to a measure preserving bijection $\mathbb{R} \to \mathbb{R},$ so I'm not sure what the limit should be. Without knowing what the limit should be, how do I prove convergence? Perhaps we can show $\{f_n\}$ is Cauchy instead and use the completeness of $L^1,$ but I'm not sure how to do that either.
We can't prove convergence directly since $f_{2n} = \chi_{[1/2, 1]}, f_{2n+1} = \chi_{[0,1/2]}$ is a counterexample, and I only need a subsequence. Any ideas?
The following sequence is a counterexample: $f_n = n \chi_{[0,1/n]}$.
Let $U$ be measurable. Since $$ \lambda( f_n^{-1}(U)) =\lambda( f_n^{-1}(U \cap (-\infty,1))) + \lambda( f_n^{-1}(U \cap [1,+\infty))) $$ we can discuss the cases $U \subset (-\infty,1)$ and $U \subset [1,+\infty)$ separately.
If $U \subset (-\infty,1)$ and $0\in U$ then $\lambda( f_n^{-1}(U)) = 1-\frac1n\to 1$. If $0\not\in U$ then $\lambda( f_n^{-1}(U)) =0$.
If $U \subset [1,+\infty)$ then $\lambda( f_n^{-1}(U)) \le \frac1n \to0$.
So $\lambda( f_n^{-1}(U))$ is convergent for all measurable subsets $U$, but $(f_n)$ does not converge in $L^1$ and does not have a converging subsequence. In particular $\|f_n -f_m\|_{L^1} \ge 1$ for $m\ne n$.