Metamath Proof Explorer


Theorem itg1addlem1

Description: Decompose a preimage, which is always a disjoint union. (Contributed by Mario Carneiro, 25-Jun-2014) (Proof shortened by Mario Carneiro, 11-Dec-2016)

Ref Expression
Hypotheses itg1addlem.1 ⊢ φ → F : X ⟶ Y
itg1addlem.2 ⊢ φ → A ∈ Fin
itg1addlem.3 ⊢ φ ∧ k ∈ A → B ⊆ F -1 k
itg1addlem.4 ⊢ φ ∧ k ∈ A → B ∈ dom ⁡ vol
itg1addlem.5 ⊢ φ ∧ k ∈ A → vol ⁡ B ∈ ℝ
Assertion itg1addlem1 ⊢ φ → vol ⁡ ⋃ k ∈ A B = ∑ k ∈ A vol ⁡ B

Proof

Step Hyp Ref Expression
1 itg1addlem.1 ⊢ φ → F : X ⟶ Y
2 itg1addlem.2 ⊢ φ → A ∈ Fin
3 itg1addlem.3 ⊢ φ ∧ k ∈ A → B ⊆ F -1 k
4 itg1addlem.4 ⊢ φ ∧ k ∈ A → B ∈ dom ⁡ vol
5 itg1addlem.5 ⊢ φ ∧ k ∈ A → vol ⁡ B ∈ ℝ
6 4 5 jca ⊢ φ ∧ k ∈ A → B ∈ dom ⁡ vol ∧ vol ⁡ B ∈ ℝ
7 6 ralrimiva ⊢ φ → ∀ k ∈ A B ∈ dom ⁡ vol ∧ vol ⁡ B ∈ ℝ
8 3 adantrr ⊢ φ ∧ k ∈ A ∧ x ∈ B → B ⊆ F -1 k
9 simprr ⊢ φ ∧ k ∈ A ∧ x ∈ B → x ∈ B
10 8 9 sseldd ⊢ φ ∧ k ∈ A ∧ x ∈ B → x ∈ F -1 k
11 1 ffnd ⊢ φ → F Fn X
12 11 adantr ⊢ φ ∧ k ∈ A ∧ x ∈ B → F Fn X
13 fniniseg ⊢ F Fn X → x ∈ F -1 k ↔ x ∈ X ∧ F ⁡ x = k
14 12 13 syl ⊢ φ ∧ k ∈ A ∧ x ∈ B → x ∈ F -1 k ↔ x ∈ X ∧ F ⁡ x = k
15 10 14 mpbid ⊢ φ ∧ k ∈ A ∧ x ∈ B → x ∈ X ∧ F ⁡ x = k
16 15 simprd ⊢ φ ∧ k ∈ A ∧ x ∈ B → F ⁡ x = k
17 16 ralrimivva ⊢ φ → ∀ k ∈ A ∀ x ∈ B F ⁡ x = k
18 invdisj ⊢ ∀ k ∈ A ∀ x ∈ B F ⁡ x = k → Disj k ∈ A B
19 17 18 syl ⊢ φ → Disj k ∈ A B
20 volfiniun ⊢ A ∈ Fin ∧ ∀ k ∈ A B ∈ dom ⁡ vol ∧ vol ⁡ B ∈ ℝ ∧ Disj k ∈ A B → vol ⁡ ⋃ k ∈ A B = ∑ k ∈ A vol ⁡ B
21 2 7 19 20 syl3anc ⊢ φ → vol ⁡ ⋃ k ∈ A B = ∑ k ∈ A vol ⁡ B