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 $n\in\mathbb{N}$.

1 Answer 1


"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.


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