0

Suppose for a sequence of functions $\{ f_n\}$ we have that $|f_n | \leq g$ where $g\in L^1$. If $f_n \to f$ in measure, is it necessarily true that $| f | \leq g$?

This seems like it must be true, and intuitively makes sense. I can't seem to prove it though or come up with a counterexample. My proof thus far is: Let $\epsilon > 0$. Then $$| f| = |f - f_n + f_n | \leq | f - f_n | + |f_n| \leq g + | f - f_n |$$ where the latter term can be made $< \epsilon$ but an arbitrarily small set. But this does not show that $|f| \leq g$ as this does not hold on the entire domain.

kam
  • 302

1 Answers1

2

One has $f_{n_k} \to f$ almost everywhere and thus $|f|\leq g$, so that $f\in L^1$.

Andrew
  • 1,679
  • 1
    For people unaware of this fact, see here https://math.stackexchange.com/questions/1006091/convergence-in-measure-implies-convergence-almost-everywhere-of-a-subsequence – Severin Schraven Aug 08 '23 at 17:25
  • Makes sense. I would just clarify that $| f| \leq g$ a.e., not everywhere. But this doesn't change the fact that $f \in L^1$. – kam Aug 08 '23 at 19:22