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Politecnico di Milano , Mathematical Engineering Real and Functional Analysis – Exercises Prof. F. Punzo, G. Verzini, February 13, 2025 E1 E2 E3 Surname/Name: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Id. No. . . . . . . . . . . . . . . . . . . . . . . . . . . . . [Solutions must be written ONL Y on these sheets, under the exercise and on the back.] [Solutions can be written in English or in Italian.] Exercise 1. [5 points] Consider the set function µ : L([0, 1]) → [0, +∞), defined by µ(E) = δ0(E) + Z E ex2 dλ ∀E ∈ L([0, 1]), where L([0, 1]) denotes the Lebesgue σ-algebra on [0, 1], λ denotes the Lebesgue measure on [0, 1] and δ0 denotes the Dirac measure concentrated at x = 0. 1. Prove that µ is a positive measure on the measurable space ([0 , 1], L([0, 1]). 2. Does there exist dλ dµ? Solution. (1) µ is a positive measure defined on L([0, 1]), indeed µ(∅) = 0 and µ is σ-additive: take {Ei}i∈N ⊂ L([0, 1]) disjoint family of measurable sets, then µ [ i∈N Ei ! = δ0 [ i∈N Ei ! + Z S i∈N Ei ex2 dx = X i∈N δ0(Ei) + X i∈N Z Ei ex2 dx = X i∈N µ(Ei), where the validity of the second equality follows by the disjoint character of the family {Ei}i∈N. (2) We check if the assumptions of the Radon-Nikodym theorem are satisfied, to get the existence (and unique- ness a.e.) of the Radon-Nikodym derivative dλ dµ. Is µ σ -finite? Yes, actually µ is a finite measure µ([0, 1]) = δ0([0, 1]) + Z 1 0 ex2 dλ ≤ 1 + e. Is λ ≪ µ? Let E ∈ L([0, 1]) such that µ(E) = 0. Then, 0 /∈ E and R E ex2 dλ = 0. In particular, since f(x) = ex2 is a positive function on the unit interval [0 , 1], we get λ(E) = 0. Therefore λ ≪ µ. Hence, we can apply the Radon-Nikodym theorem to affirm that there exists dλ dµ such that λ(E) = Z E dλ dµ(x) dµ…