实变函数与泛函分析基础第四章习题答案

1.f(x)Er,E[f > r]E[f = r]f(x)r,E[f > r]α,{rn}αE[f > a] =∞n= 1E[f > rn],E[f > rn]E[f > α]f(x)Er,E[f = r]f(x)E = (−∞, ∞), z(−∞, ∞)x ∈ z, f(x) = √2,r,E[f = r] = ∅E[f > √n→∞E| fn − f |< 1k,x ∈ limk.kx ∈∞k= 1limk.x ∈∞k= 1limk,ǫ> 0,k0,1n→∞E| fn − f |<1k0,| fn(x) − f(x) |<1n→∞E| fn − f |< 1n→∞fn(x)n→∞fn = +∞]fn+∞E[limn→∞fn > limn→∞fn = +∞] − E[limn→∞fn > limE[ limlimn→∞fn = −∞] ∪ E[n→∞fn]4.E[0, 1]f(x) =x,x ∈ E,−x,x ∈ [0, 1] − E.f(x)[0, 1]| f(x) |f(x)0 ∈ E,E[f ≥ 0] = E0 ∈ E,E[f > 0] = Ef(x)x ∈ [0, 1]| f(x) |= x| f(x) |[0, 1]5.fn(x)(n = 1, 2, · · · )Ea.e.| fn |a.e.f.ǫ > 0cE0 ⊂ E, m(E\E0) < ǫ,E0n| fn(x) |≤ c.mE < ∞.E[| fn |= ∞], E[fn → f]n = 0, 1, 2, · · · .E1 = E[fn →f]∪(∞n= 0E[| fn |= ∞]),mE1 = 0.E −E1fn(x)f(x).E2 = E −E1,x ∈ E2, supn| fn(x) |< ∞.E2 =∞k= 1E2[supn| fn |≤ k], E2[supn| fn |≤ k] ⊂ E2[supn| fn |≤ k + 1].mE2 =limk→∞mE2[supn| fn |≤ k].k0mE2 − mE2[supn| fn |≤ k0] < ǫ.E0 = E2[supn| fn |≤ k0], c = k0.E0n, | fn(x) |≤ c,m(E − E0) = m(E − E2) + m(E2 − E0) < ǫ.6.f(x)(−∞, ∞)g(x)[a, b]f(g(x))E1 = (−∞, ∞), E2 = [a, b].f(x)E1c,E1[f < c]E1[f > c] =∞n= 1(αn, βn),(αn, βn)(αn−∞,βn+∞).E2[f(g) > c] =∞n= 1E2[αn < g < βn] =∞n= 1(E2[g > αn] ∩E2[g < βn]),gE2E2[g > αn], E2[g < βn]E[f(g) > c]7.fn(x), (n = 1, 2, · · ·)E””f(x),{fn}a.e.f.fn(x)E””f(x),δ > 0,Eδ ⊂ E,m(E − Eδ) < δfnEδf(x).E0Efnδ, E0 ⊂ E − Eδ(Eδfn),mE0 ≤ m(E − E0) < δ,δ → 0,mE0 = 0.fn(x)Ea.e.f(x)().28.f(x)Eδ > 0,Eδ ⊂ Ef(x)Eδm(E − Eδ) < δ,f(x)Ea.e.1/n,En ⊂ E,f(x)Enm(E − En) < 1n.n → ∞,mE0 = 0. E = (E − E0) ∪ E0 = (∞n= 1En) ∪ E0 ...