isabelle


set integrable with functions multiplication


I'm trying to prove this lemma:
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)"
and
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)"
but I couldn't. I've seen that it is proved for addition and subtraction:
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)
or even for scalar multiplication but not for two functions multiplication?
The problem is that it is quite simply not true. The function f(x) = 1 / sqrt(x) is integrable on the set (0;1], and the integral has the value 2. Its square f(x)² = 1 / x on the other hand is not integrable on the set (0;1]. The integral diverges.

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