MATH 202 B - Problem Set 2
نویسنده
چکیده
we have by construction μ(En) = an. An important observation is that a subset of consecutive intervals En, En+1, . . . , En+k are either pairwise disjoint, or they cover the whole interval [0, 1]. Now for all n, let fn = 1En . Then we have • ‖fn‖1 = ∫ X |1En |dμ = μ(En) = an. • for all x ∈ [0, 1], (fn(x))n does not converge. Indeed, if we suppose by contradiction that (fn(x))n converges for some x, then since every fn takes values in {0, 1}, we have lim fn(x) is either 0 or 1. We show that both cases lead to a contradiction
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MATH 202 B - Problem Set 10
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