Does the Komlós theorem hold in infinite measure spaces?

Question

I read an article, and they use a certain theorem, called Komlós theorem, which says:

Let $(E,\mathcal {A}, \mu ) $ be a finite measure space and $ (f_n)_{n\geq 1} \subset \mathcal {L}_{\mathbb {R}}^1$ is a sub-sequence with : $$\sup_n \int_{E}{|f_n| d\mu} < \infty .$$Then there exist $ h _{\infty} \in \mathcal {L}_{\mathbb {R}}^1 $ and a sub-sequence $ (g_k)_k $ of $(f_n)_n $ such that for every sub-sequence $ (h_m)_m $ of $(g_k)_k$ : $$ \frac{1}{i}\sum_{j=1}^{i}{h_j}\to h _{\infty} \text{ a.s. }$$

if the space is of infinite measure, is that this result rest valid ?

please an idea?

Answer 1

Yes, it is also true for general measure spaces. This was observed by Chatterji in his paper A general strong law but it seems that this generalization was pretty obvious to him, and so he didn't give a (detailed) proof how the general case follows from the finite measure case. Here is one possible approach:

Step 1: Komlós theorem is true for $\sigma$-finite measure spaces.

Let $(E,\mathcal{A},\mu)$ be a $\sigma$-finite measure spaces and $(f_n)_{n \in \mathbb{N}} \subseteq L^1(E)$ such that $M=\sup_n \|f_n\|_{L^1}< \infty$. Since the measure space is $\sigma$-finite, there is a sequence $(E_n)_n \subseteq \mathcal{A}$ such that $E_n \uparrow E$ and $\mu(E_n)<\infty$ for all $n$. Without loss of generality, $\mu(E_n)>0$ for all $n$. If we define $$u(x) := \sum_{n \in \mathbb{N}} \frac{1}{n^2 \mu(E_n)} 1_{E_n}(x),$$ then $u \geq 0$ is integrable with respect to $\mu$. In particular, $$\nu(A) := \int_A u(x) \, \mu(dx), \qquad A \in \mathcal{A}$$ defines a finite measure on $(E,\mathcal{A})$. Since $u$ is bounded, the sequence $(f_n)_{n \in \mathbb{N}}$ satisfies $\sup_{n \in \mathbb{N}} \int |f_n| \, d\nu< \infty$. Consequently, by Komlós theorem (for finite measures), there exist $h_{\infty} \in L^1(\nu)$ and a subsequence $(g_k)_k$ of $(f_n)_n$ such that any subsequence $(h_i)_i$ of $(g_k)_k$ satisfies

$$\frac{1}{i} \sum_{j=1}^i h_j \to h_{\infty}$$

almost surely with respect to $\nu$. Since the density $u$ is strictly positive, the convergence holds also almost surely with respect to $\mu$. From Fatou's lemma, we get

\begin{align*} \int |h_{\infty}| \, d\mu \leq \liminf_{i \to \infty} \frac{1}{i} \sum_{j=1}^i \underbrace{\int |h_j| \, d\mu}_{\leq M < \infty} \leq M, \end{align*}

and so $h_{\infty} \in L^1(\mu)$. This finishes the proof for the $\sigma$-finite case.

Step 2: Komlós theorem is true for general measure spaces.

We can boil down everything to the case that we have a $\sigma$-finite measure spaces. Fix an arbitrary measure space $(E,\mathcal{A},\mu)$ and a sequence of integrable functions $f_n$ with $\sup_n \|f_n\|_{L^1}< \infty$. Set

$$E_n := \left\{x \in E; \max \{|f_1(x)|,\ldots,|f_n(x)|\}> \frac{1}{n} \right\}, \quad n \geq 1,$$

then, by definition, the sets $E_n$ are increasing in $n$. Define $\tilde{E} := \bigcup_{n \in \mathbb{N}} E_n$, $\tilde{\mathcal{A}} := \mathcal{A} \cap \tilde{E}$ (trace $\sigma$-algebra) and $\tilde{\mu} := \mu|_{\tilde{E}}$ (trace measure), then $(\tilde{E},\tilde{\mathcal{A}},\tilde{\mu})$ is a $\sigma$-finite measure spaces. By Step 1, we can find $h_{\infty} \in L^1(\tilde{E})$ and a subsequence of $(g_k)_k$ of $(f_n)_n$ such that every subsequence $(h_m)_m$ of $(g_k)_k$ satisfies $$\frac{1}{i} \sum_{j=1}^i h_j \to h_{\infty}$$ almost surely with respect to $\tilde{\mu}=\mu|_{\tilde{E}}$, i.e.

$$\mu \left\{ x \in \tilde{E}; \lim_{i \to \infty} \frac{1}{i} \sum_{j=1}^i h_j(x) \neq h_{\infty}(x) \right\}=0.$$

On the other hand, if $x \in E \backslash \tilde{E}$, then $f_n(x)=0$ for all $n \in \mathbb{N}$ and so, trivially,

$$\lim_{i \to \infty} \frac{1}{i} \sum_{j=1}^i h_j(x) =0$$

for any such $x$ and any subsequence of $(f_n)_n$. Consequently, if we extend $h_{\infty}$ to $E$ by setting $h_{\infty}(x)=0$ for all $x \in E \backslash \tilde{E}$, then we get

$$\mu \left\{ x \in E; \lim_{i \to \infty} \frac{1}{i} \sum_{j=1}^i h_j(x) \neq h_{\infty}(x) \right\}=0.$$