Dose convergence in measure imply convergence almost every where?

Question

I learned that f_{n} converge f in measure imply the existence of subsequnce n_{j} that f_{n_{j}} converge f almost everywhere. Then is it true that if subsequnce of the f_{n} converges almost everywhere to g , then f=g ??

Answer 1

Yes, it is true (almost everywhere). Suppose $f_{n_j} \to g$ almost everywhere, as subsequences of convergent sequence are convergent, we have $f_{n_j} \to f$ in measure. By the cited theorem, for some subsequence, $f_{n_{j_k}} \to f$ almost everywhere. As subsequence of $(f_{n_j})$, we still have $f_{n_{j_k}} \to g$ almost everywhere. By uniqueness of limits, $f =g $ a. e.