### Decomposition of max

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128 A Appendices

Hence, the preceding argument (using a dense sequence) applied for n (F (x)) n1 guarantees the measurability of x k0 k (F (x)) (for each n) and we are done. Let be the symmetric norm for (nite) sequences corresponding to . Since F (x) lim (0 (F (x)), 1 (F (x)), . . . , n (F (x)), 0, 0, . . . ), n

to prove the lemma it suces to check the measurability of x (0 (F (x)), 1 (F (x)), . . . , n (F (x)), 0, 0, . . . ) [0, ) for each xed n. Note that this map is the composition of the measurable map x (0 (F (x)), 1 (F (x)), . . . , n (F (x))) (thanks to the rst half of the proof) followed by (a0 ,...

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