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I'm reading a book on Lyapunov Exponents by Lian and Lu in which they refer to strong measurability of operator-valued maps. They define this by saying an operator valued map $T:\Omega\to L(X,X)$ is strongly measurable if for any fixed $x\in X$, the map $\omega\mapsto T(\omega)(x)$ is measurable (without any mention of the $\sigma$-algebra that they're referring to).

I looked on Wikipedia, but wasn't able to get much more there, nor was I able to find a solid reference. I think I now understand what the authors meant and have proved a few lemmas solidifying the definition etc., but I would like to be able to look at a more solid reference to avoid re-inventing the wheel if possible.

Thanks for any suggestions...

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The topology on $X$ required here is the norm topology of a Banach space. This definition is analogous to the "strong operator topology" which you probably can find.

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So that is, for fixed $x \in X$, we have $\omega \mapsto T(\omega)(x)$ as a map from $\Omega$ to $X$. The condition is then that this map should be measurable with respect to some given $\sigma$-algebra on $\Omega$, and the Borel $\sigma$-algebra generated by the norm topology on $X$. –  Nate Eldredge Jan 7 '11 at 15:21
    
Thanks for this. What I decided it had to mean was that the sigma algebra on $X$ had to be the Borel $\sigma$-algebra of the norm topology on $X$. In the case in question, both $X$ and its dual are separable. In this case I think that strongly measurable is equivalent to being measurable with respect to the Borel $\sigma$-algebra of the strong operator topology on $L(X)$. It would sure be nice to have a reference though... –  Anthony Quas Jan 7 '11 at 19:36
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