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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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  • $\begingroup$ May I know whether you found a reference of this result eventually? Does the claim hold for any Banach space $X$? $\endgroup$
    – John
    May 10, 2022 at 6:51
  • $\begingroup$ In the end, I re-invented the wheel. There is an appendix to my <a href="A semi-invertible operator Oseledets theorem">paper</a> with Cecilia González-Tokman, where the definition is spelled out and some basic lemmas are proved. $\endgroup$ May 11, 2022 at 7:09

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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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  • $\begingroup$ 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 $X$. $\endgroup$ Jan 7, 2011 at 15:21
  • $\begingroup$ 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... $\endgroup$ Jan 7, 2011 at 19:36

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