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 som...