Is it true that $\min\{h, f_n\}\to h$ in measure provided $f_n\to f$ in measure?

Question

Let $E\subseteq \mathbb{R}^n$ be a measurable set with finite measure. $\{f_n\}$ and $f$ are non-negative measurable functions on $E$, s.t. $f_n\to f$ in measure on $E$. If $h$ is a measurable, non-negative, bounded function on $E$ s.t. $0\le h(x)\le f(x)$ for all $x\in E$, is it true that $\min\{h, f_n\}\to h$ in measure on $E$?

Motivation for this question: I want to show that Fatou's Lemma is still true if the hypothsis "a.e. convergence" is replaced by "convergence in measure". I know a classical proof is to assume the contrary. But I wonder if it can be proved by modifying the proof for "a.e. convergence". But when repeating the proof I have to prove the above claim.

Answer 1

Yes the claim is correct. Let $\epsilon >0$

$$h(x) - \min\{h(x), f_n(x)\} > \epsilon \; \implies \; f_n(x) < h(x) - \epsilon\le f(x) - \epsilon\; \implies f(x) - f_n(x) > \epsilon.$$

Can you finish the proof?

Answer 2

The claim is correct.

Lemma: $f_n \to f$ in measure as $n \to \infty$ if and only if, for every given subsequence $\{n_k\}_{k=1}^{\infty}$, there is a subsubsequence $\{n_{k_l}\}_{l=1}^{\infty}$ such that $f_{n_{k_l}} \to f$ in measure.

The lemma is immediately proved by contradiction.

Let a subsequence $\{n_k\}_{k=1}^{\infty}$ be given. We know that $f_{n_k} \to f$ in measure as $k \to \infty$. Choose subsubsequence $\{n_{k_l}\}_{l=1}^{\infty}$ such that $f_{n_{k_l}} \to f$ almost surely as $l \to \infty$. We see that, almost surely (and in particular in measure), $\min(h,f_{n_{k_l}}) \to h$ as $l \to \infty$. Therefore, $\min(h,f_{n}) \to h$ in measure as $n \to \infty$.