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Let $A$ be a $\sigma$-algebra. A family $\{A_{m,n}\}$ of elements of $A$ satisfies the infinite distributive law in case

$$\bigwedge_{m\in\omega}\bigvee_{n\in\omega} A_{m,n}=\bigvee_{\alpha\in\omega^\omega}\bigwedge_{m\in\omega}A_{m,\alpha(m)}$$

where $\omega^\omega$ denotes the set of all mappings of $\omega$ into $\omega$. Further, in every $\sigma$-algebra, the following inequality always hold:

$$\bigwedge_{m\in\omega}\bigvee_{n\in\omega} A_{m,n}\geq\bigvee_{\alpha\in\omega^\omega}\bigwedge_{m\in\omega}A_{m,\alpha(m)}$$

Where can I find a proof of this inequality?

Ben
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2 Answers2

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Fix $\alpha \in \omega^{\omega}$.
For every $m \in \omega$, $$A_{m,\alpha(m)} \leq \bigvee_{n\in\omega}A_{m,n}$$ because $\alpha(m) \in \omega$, and so $A_{m,\alpha(m)}$ is one of the joinands. Thus, $$\bigwedge_{m\in\omega} A_{m,\alpha(m)} \leq \bigwedge_{m \in \omega} \bigvee_{n\in\omega}A_{m,n},$$ and the desired inequality follows.

Notice that this generalizes the well-known lattice distributive inequality: $$(x \wedge y) \vee (x \wedge z) \leq x \wedge (y \vee z).$$

amrsa
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The statement $$\bigvee\limits_\alpha \bigwedge\limits_m A_{m, \alpha(m)} \leq \bigwedge\limits_m \bigvee\limits_n A_{m, n}$$

is equivalent to the statement that for all $w$ and for all $\alpha$,

$$\bigwedge\limits_m A_{m, \alpha(m)} \leq \bigvee\limits_n A_{w, n}$$

In particular, by the definitions of $\land$ and $\lor$, we have

$$\bigwedge\limits_m A_{m, \alpha(m)} \leq A_{w, \alpha(w)} \leq \bigvee\limits_n A_{w, n}$$

which completes the proof.

Mark Saving
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