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Let $X \neq \varnothing$ be a set, $\{x_k\}_{k\in \Bbb{N}} \subset X$ and $\{f_n : X \hookrightarrow[0,\infty]\}_{n\in \Bbb{N}}$ such that $\forall n \in \Bbb{N} \, \forall x \in X : f_n(x) \leq f_{n+1}(x)$.

I want to know in which conditions $$\sum_{k\in \Bbb{N}} \lim_{m \rightarrow \infty} f_m(x_k) = \lim_{m \rightarrow \infty} \sum_{k\in \Bbb{N}} f_m(x_k)$$

My attempt guides me to the following by knowing that the limit of a increasing sequence is greater than all the elements of the sequence: $$\forall n \in \Bbb{N} \, \forall x \in X : f_n(x) \leq f_{n+1}(x) \Rightarrow \forall n,k \in \Bbb{N} : f_n(x_k) \leq f_{n+1}(x_k) \Rightarrow \forall n,k \in \Bbb{N}:f_n(x_k) \leq \lim_{m \rightarrow \infty} f_m(x_k) \Rightarrow \forall n \in \Bbb{N}:\sum_{k\in \Bbb{N}} f_n(x_k) \leq \sum_{k\in \Bbb{N}} \lim_{m \rightarrow \infty} f_m(x_k) \Rightarrow \lim_{m \rightarrow \infty} \sum_{k\in \Bbb{N}} f_m(x_k) \leq \sum_{k\in \Bbb{N}} \lim_{m \rightarrow \infty} f_m(x_k) $$

I don't know if the other inequality is true, neither which conditions should make it true. Any possible answers would be appreciated.

Angelo
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Superdivinidad
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1 Answers1

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I apologize for my previous mistakes. I'm also currently learning real analysis and my knowledge is quite fragmented.

For simplicity I will write $f_m(k)$ instead of $f_m(x_k)$.Tannery's theorem in my previous response does not make full use of the conditions of monotonicity and non-negativeness of $f_m(x)$. Instead, this is exactly the statement of Monotone convergence theorem:

If for all $m,k\in\mathbb N, f_m(k)\in[0,+\infty]$ and $f_m(k)\le f_{m+1}(k)$, then $$\sum_{k\in \mathbb N}\lim_{m\to\infty}f_m(k)=\lim_{m\to\infty}\sum_{k\in \mathbb N}f_m(k)$$

Ricky
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    This is a related theorem ... not as powerful as the Dominated Convergence Theorem, but provides sufficient conditions for interchanging limits – Mark Viola Sep 09 '23 at 16:58
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    Ricky, in the case of non-negative functions and monotone convergence, the hypothesis $\sum_{k\in\Bbb N}M_k<\infty$ is not necessary, so you can remove it. You could also remove the hypothesis $\lim\limits_{m\to\infty}f_m(k)\leqslant M_k$. See Monotone Convergence Theorem. – Angelo Sep 10 '23 at 04:37
  • Sorry for my previous mistakes...My knowledge of real analysis is quite fragmented and makes me ignorant of such famous theorems as MCT. I suppose I had better learn more before answering... – Ricky Sep 10 '23 at 06:00