# How to characterize the order convergence in Bochner-integrable functions space?

Let $$(\Omega,\Sigma,\mu)$$ a finite measure space. We want to characterize the order convergence (for sequences) in Bochner integrable functions space $$L^1(\mu,X)$$, $$X$$ Banach lattice.

In $$L^p$$ we have: A sequence $$(f_n)_1^\infty\subset L^P(\mu)$$ is order convergent to $$f$$ as $$n\to\infty$$ if and only if there exists some $$0\leq g\in L^p(\mu)$$ such that $$|f_n|\leq g$$ a.e. and $$f_n\to f$$ a.e.

We want a similar result for $$L^1(\mu,X)$$: If $$(f_n)_1^\infty\subset L^1(\mu,X)$$ is a sequence of functions, with $$X$$ $$\sigma$$-order continuous. Then $$(f_n)$$ is order convergent to $$f$$ as $$n\to\infty$$ if and only if there exists some $$0\leq g\in L^1(\mu,X)$$ such that $$|f_n|\leq g$$ a.e. and $$f_n\to f$$ a.e.

This is my try:

$$\Rightarrow]$$ Suppose $$f_n\xrightarrow{o} f$$. Then, exists $$g_n\downarrow 0$$ such that $$|f_n-f|\leq g_n$$ for all $$n\in\mathbb{N}$$. Since $$X$$ is $$\sigma$$-order continuous, $$g_n\downarrow 0$$ implies $$\|g_n\|\downarrow 0$$ a.e., then $$\|g_n\|\to 0$$ a.e. Hence $$g_n\to 0$$ a.e. Thus $$f_n\to f$$ a.e. On the other hand, note that $$|f_n|=|f+f_n-f|\leq |f|+|f_n-f|\leq |f|+g_n\leq |f|+g_1$$. Define $$g=|f|+g_1\geq 0$$ we done.

$$\Leftarrow]$$ Suppose, without loss of generality, $$f_n\to 0$$ a.e. and there exists $$g\in L(\mu,X)$$ such that $$|f_n|\leq g$$. Define $$g_n=\sup\{|f_m|:m\geq n\}$$ for each $$n\in\mathbb{N}$$. Note that $$g_n\downarrow$$. Since $$X$$ is $$\sigma$$-order continuous, $$L^1(\mu,X)$$ is $$\sigma$$-Dedekind complete, so $$g_n\in L^1(\mu,X)$$ for every $$n\in\mathbb{N}$$.

If $$f_n\to 0$$ a.e., then $$g_n\to 0$$ a.e.? Is this true? (*)

Since $$g_n\leq g$$ and $$g_n\to 0$$ a.e., by the monotone convergence theorem, $$g_n\to 0$$. Thus $$f_n\xrightarrow{o}0$$.

Some notes and comments:

• First, (*) is true? I have not been able to prove it.
• My proof is wrong? Some alternative or new ideas?
• If $$X$$ is $$\sigma$$-order continuous, then $$L^1(\mu,X)$$ is $$\sigma$$-Dedekind complete. This is a proven fact.

Notation, definitions, etc:

• $$L^1(\mu,X)=\{f:\Omega\to X \, |\, f \text{ is Bochner integrable}\}.$$
• $$g_n\downarrow 0$$ means $$g_n\leq g_{n+1}$$ for all $$n\in\mathbb{N}$$ and $$\inf_n g_n=0$$.
• $$X$$ is $$\sigma$$-order continuous if $$x_n\downarrow 0$$ implies $$\|x_n\|\downarrow 0$$.
• $$X$$ is $$\sigma$$-Dedekind complete if $$(x_n)\subset X$$ is a bounded sequence implies $$\sup_nx_n,\inf_nx_n\in X$$.
• $$X$$ Banach lattice, for all $$x\in X$$ we define $$|x|=\sup\{-x,x\}$$,.
• In a Riesz space, a sequence $$(x_n)\subset X$$ is called order convergent to $$x$$ if there exists a sequence $$(y_n)\subset X$$ such that $$y_n\downarrow 0$$ and $$|x_n-x|\leq y_n$$ for all $$n\in\mathbb{N}$$.

"If $$f_n\to 0$$ a.e., then $$g_n\to 0$$ a.e.? Is this true? (*)"
No, this is not true in general. E.g., suppose that (i) $$\mu$$ is the only probability measure over the singleton set $$\Omega:=\{0\}$$, (ii) $$X:=L^1[0,1]$$, and (iii) $$f_n(0):=1_{I_n}\in X$$ (so that $$f_n\in L^1(\mu,X)$$), where $$(I_n)$$ is any sequence of subintervals of the interval $$[0,1]$$ such that the length $$|I_n|$$ of $$I_n$$ goes to $$0$$ (as $$n\to\infty$$) and $$\bigcup\limits_{m\colon\, m\ge n}I_m=[0,1]$$ for all natural $$n$$.
Then $$\|f_n(0)\|_X=|I_n|\to0$$ and hence $$f_n(0)\to0$$ and $$f_n\to0$$ $$\mu$$-a.e. However, $$g_n=\sup\{|f_m|\colon m\ge n\}=1$$ for all $$n$$ and hence $$g_n\not\to0$$ $$\mu$$-a.e.