Question

我正试图证明这个问题：

  lemma set_integral_mult:
  fixes f g :: "_ ⇒ _ :: {banach, second_countable_topology}"
  assumes "set_integrable M A (λx. f x)"  "set_integrable M A (λx. g x)"
    shows "set_integrable M A (λx. f x * g x)"

和

  lemma set_integral_mult1:
  fixes f :: "_ ⇒ _ :: {banach, second_countable_topology}"
  assumes "set_integrable M A (λx. f x)"  
    shows "set_integrable M A (λx. f x * f x)"

但我不能。我已经看到它被证明了加法和减法：

  lemma set_integral_add [simp, intro]:
  fixes f g :: "_ ⇒ _ :: {banach, second_countable_topology}"
  assumes "set_integrable M A f" "set_integrable M A g"
  shows "set_integrable M A (λx. f x + g x)"
    and "LINT x:A|M. f x + g x = (LINT x:A|M. f x) + (LINT x:A|M. g x)"
  using assms by (simp_all add: scaleR_add_right)

  lemma set_integral_diff [simp, intro]:
  assumes "set_integrable M A f" "set_integrable M A g"
  shows "set_integrable M A (λx. f x - g x)" and "LINT x:A|M. f x - g x =
    (LINT x:A|M. f x) - (LINT x:A|M. g x)"
  using assms by (simp_all add: scaleR_diff_right)

甚至是标量乘法而不是两个函数乘法？

Answer 1

问题在于它根本不是真的。函数f(x) = 1 / sqrt(x)在集合（0; 1）上是可积的，而积分的值是2.另一方面，它的平方f(x)² = 1 / x在集合（0; 1）上是不可积的。积分分歧。

设置可积函数乘法

1 个答案: