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Let $(\Omega, \mathcal F, \mu)$ be a measure space with $\mu(\Omega) < \infty$. Let $(f_n) \subset L^1(\Omega)$ and $f:\Omega \to \mathbb R$ be measurable such that $f_n \to f$ $\mu$-a.e. Let's call

  • Assumption 1: there is $r>0$ such that $$ \int |f_n| \le r \quad \forall n \in \mathbb N. $$

  • Assumption 2: $f_n \le f_{n+1}$ for all $n$ and there is $r>0$ such that $$ \int f_n \le r \quad \forall n \in \mathbb N. $$

If Assumption 1 holds, then $f \in L^1 (\Omega)$ by Fatou's lemma.

Does $f \in L^1 (\Omega)$ hold when we replace Assumption 1 with Assumption 2?

Akira
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    (1) follows from Fatou's Lemma: $\int |f|,d\mu\leq \liminf_n\int |f_n|,d\mu\leq r$.(2) is a straight application of monotone convergence: $0\leq F_n=f_n-f_1\nearrow f-f_1$ – Mittens Apr 03 '23 at 18:39
  • @OliverDíaz doesn't monotone convergence require non-negativity of the sequence? – fwd Apr 03 '23 at 18:41
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    @fwd Replace $f_n$ with $f_n-f_1.$ – Ryszard Szwarc Apr 03 '23 at 18:41
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    @fwd: If all the elements of the sequence are integrable, a simple trick (see the edit to my comment) does the trick. – Mittens Apr 03 '23 at 18:42
  • @OliverDíaz Could you post your comment as an answer? – Akira Apr 03 '23 at 18:44
  • @Akira: just did as a community wiki. This question did not require too much in my opinion, but now you have an answer for your purposes. Regards! – Mittens Apr 03 '23 at 18:48
  • @OliverDíaz I have to accept an answer that solved my question :v – Akira Apr 03 '23 at 18:48

1 Answers1

2

(1) follows from Fatou's Lemma: $\int |f|\,d\mu\leq \liminf_n\int |f_n|\,d\mu\leq r$.

(2) is a straight application of monotone convergence: $0\leq F_n=f_n-f_1\nearrow f-f_1$

Mittens
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