[32] By Exercise 4.5.4, Mnorm dominates M.

(a) Show that M does not multiplicatively dominate Mnorm.

(b) Show that for each normalizer a defining the measure M

(x) =

a(x)M(x) we have M(ω1:n) = o(M

(ω1:n), for some infinite ω.

(c) (Open) Item

(b) with ‘all’ substituted for ‘some.’

Comments. Item

(b) implies that M does not dominate any of its normalized versions M

. This is a special case of Exercise 4.5.6.