Metamath Proof Explorer


Theorem itg1le

Description: If one simple function dominates another, then the integral of the larger is also larger. (Contributed by Mario Carneiro, 28-Jun-2014) (Revised by Mario Carneiro, 6-Aug-2014)

Ref Expression
Assertion itg1le ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 ∧ F ≤ f G → ∫ 1 ⁡ F ≤ ∫ 1 ⁡ G

Proof

Step Hyp Ref Expression
1 simp1 ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 ∧ F ≤ f G → F ∈ dom ⁡ ∫ 1
2 0ss ⊢ ∅ ⊆ ℝ
3 2 a1i ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 ∧ F ≤ f G → ∅ ⊆ ℝ
4 ovol0 ⊢ vol * ⁡ ∅ = 0
5 4 a1i ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 ∧ F ≤ f G → vol * ⁡ ∅ = 0
6 simp2 ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 ∧ F ≤ f G → G ∈ dom ⁡ ∫ 1
7 simpl ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 → F ∈ dom ⁡ ∫ 1
8 i1ff ⊢ F ∈ dom ⁡ ∫ 1 → F : ℝ ⟶ ℝ
9 ffn ⊢ F : ℝ ⟶ ℝ → F Fn ℝ
10 7 8 9 3syl ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 → F Fn ℝ
11 simpr ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 → G ∈ dom ⁡ ∫ 1
12 i1ff ⊢ G ∈ dom ⁡ ∫ 1 → G : ℝ ⟶ ℝ
13 ffn ⊢ G : ℝ ⟶ ℝ → G Fn ℝ
14 11 12 13 3syl ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 → G Fn ℝ
15 reex ⊢ ℝ ∈ V
16 15 a1i ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 → ℝ ∈ V
17 inidm ⊢ ℝ ∩ ℝ = ℝ
18 eqidd ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 ∧ x ∈ ℝ → F ⁡ x = F ⁡ x
19 eqidd ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 ∧ x ∈ ℝ → G ⁡ x = G ⁡ x
20 10 14 16 16 17 18 19 ofrval ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 ∧ F ≤ f G ∧ x ∈ ℝ → F ⁡ x ≤ G ⁡ x
21 20 3exp ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 → F ≤ f G → x ∈ ℝ → F ⁡ x ≤ G ⁡ x
22 21 3impia ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 ∧ F ≤ f G → x ∈ ℝ → F ⁡ x ≤ G ⁡ x
23 eldifi ⊢ x ∈ ℝ ∖ ∅ → x ∈ ℝ
24 22 23 impel ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 ∧ F ≤ f G ∧ x ∈ ℝ ∖ ∅ → F ⁡ x ≤ G ⁡ x
25 1 3 5 6 24 itg1lea ⊢ F ∈ dom ⁡ ∫ 1 ∧ G ∈ dom ⁡ ∫ 1 ∧ F ≤ f G → ∫ 1 ⁡ F ≤ ∫ 1 ⁡ G