Use Theorem 24.6 to show that the martingale of Example 25.14 is uniformly integrable.

Data from example 25.14

Data from theorem 24.6

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Example 25.14 Consider in R" the half-open squares Qk (2):=z+[0,2 k)", kez, ze 2-kz", with lower left corner z and side-length 2-k. Then A=0 (Qk (2): z 2-kZ"), kZ, Ak defines a (two-sided infinite) filtration ...calo cao cao cap cao ch of sub-o-algebras of B(R"). The superscript [0]' indicates that the square lat- tice in each contains some square with the origin 0 ER" as lower left corner. Just as in Example 23.3(ix), one sees that for a function f= L (X") 1 Su(x) == V(Qt(e)) ( SdX" 1 Qu(z)(x), KEZ, X" Z2-k7h is a martingale - if you are unhappy about the two-sided infinite index set, then think of (fk,ken, as a martingale and of (f-k, kEN martingale. For the square maximal function fo] (x):=sup f(x) = sup KEZ Se X" (Q) (25.13) k) as a backwards \dx"