Does convergence in measure imply convergence in mean?

Question

Here is the proposition. I doubt if the author's proof is correct.

$\mathscr{L}^{1}(X,\mathscr{A},\mu,\mathbb{R})$ is the set of all real-valued ([rather than $[-\infty,+\infty]$] -valued) integrable functions on $X$.

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The first part of the proof is easy. But I doubt whether the second part is correct. That is to say, when these assumptions are satisfied, does convergence in measure imply convergence in mean?

The red line part is confusing. In Royden's Real Analysis, the red line part is true when the assumption $\mu(X)<+\infty$ is added. See Corollary 19 below. However, Cohn doesn't give this assumption.

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For the second part of proof, is Cohn's approach correct? Do we need the assumption that the measure is finite?

Answer 1

In any measure space if a sequence $\{f_n\}$ converges in measure then there is a subsequence which converges almost everywhere. The given measure need not be finite. Proof: choose $k_n$ such that $k_n <k_{n+1}$ and $\mu \{x:|f_{k_n}(x)-f(x)| >\frac 1 {2^{n}}\} <\frac 1 {2^{n}}$. Then $\sum_n \mu\{x:|f_{k_n}(x)-f(x)| >\frac 1 {2^{n}}\} <\infty$. This implies that $\mu (\lim \sup \{x:|f_{k_n}(x)-f(x)| >\frac 1 {2^{n}}\}=0$. it follows that $f_{k_n} \to f$ almost everywhere.

Answer 2

The part underlined in red does not require the measure space to be finite. It is mostly a consequence of the Borel-Cantelli lemma, which works in any measure space. It's a bit vague, but I always remember it as "if a sequence of functions converges in measure fast enough, then it converges almost everywhere" and taking subsequences lets you increase the rate of convergence.