黎曼引理的勒贝格积分推广
题目在下图中,大家请看

任意ε>0,
1.
有K>0,使∫_{(K,+∞)}|f(x)|dx+∫_{(0,1/K)}|f(x)|dx≤ε

2.
又有函数g(x)=∑_{1≤s≤S}c(s)E_{[d(s-1),d(s))(x),
其中E_{[d(s-1),d(s))(x)为[d(s-1),d(s))的特征函数,
1/K=d(0)<...<K=d(S),满足:
∫_{[1/K,K]}|f(x)-g(x)|dx≤ε

3.
==>
|∫_{[a(λ),b(λ)]}f(x)cos(λx)dx-
-∫_{[1/K,K]∩[a(λ),b(λ)]}g(x)cos(λx)dx|≤
≤∫_{(K,+∞)∩[a(λ),b(λ)]}|f(x)cos(λx)|dx+
+∫_{(0,1/K)∩[a(λ),b(λ)]}|f(x)cos(λx)|dx+
+∫_{[1/K,K]∩[a(λ),b(λ)]}|[f(x)-g(x)]cos(λx)|dx≤2ε

4.
∫_{[1/K,K]∩[a(λ),b(λ)]}g(x)cos(λx)dx=
=∑_{1≤s≤S}c(s)∫_{[d(s-1),d(s)∩[a(λ),b(λ)]}cos(λx)dx
==>
M>0,当λ>M
|∫_{[1/K,K]∩[a(λ),b(λ)]}g(x)cos(λx)dx|≤ε
==>
|∫_{[a(λ),b(λ)]}f(x)cos(λx)dx|≤3ε
==>
Lim_{λ→+∞}∫_{[a(λ),b(λ)]}f(x)cos(λx)dx=0.