Metamath Proof Explorer


Theorem itg1ge0

Description: Closure of the integral on positive simple functions. (Contributed by Mario Carneiro, 19-Jun-2014)

Ref Expression
Assertion itg1ge0 ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F → 0 ≤ ∫ 1 ⁡ F

Proof

Step Hyp Ref Expression
1 i1frn ⊢ F ∈ dom ⁡ ∫ 1 → ran ⁡ F ∈ Fin
2 difss ⊢ ran ⁡ F ∖ 0 ⊆ ran ⁡ F
3 ssfi ⊢ ran ⁡ F ∈ Fin ∧ ran ⁡ F ∖ 0 ⊆ ran ⁡ F → ran ⁡ F ∖ 0 ∈ Fin
4 1 2 3 sylancl ⊢ F ∈ dom ⁡ ∫ 1 → ran ⁡ F ∖ 0 ∈ Fin
5 4 adantr ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F → ran ⁡ F ∖ 0 ∈ Fin
6 i1ff ⊢ F ∈ dom ⁡ ∫ 1 → F : ℝ ⟶ ℝ
7 6 adantr ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F → F : ℝ ⟶ ℝ
8 7 frnd ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F → ran ⁡ F ⊆ ℝ
9 8 ssdifssd ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F → ran ⁡ F ∖ 0 ⊆ ℝ
10 9 sselda ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F ∧ x ∈ ran ⁡ F ∖ 0 → x ∈ ℝ
11 i1fima2sn ⊢ F ∈ dom ⁡ ∫ 1 ∧ x ∈ ran ⁡ F ∖ 0 → vol ⁡ F -1 x ∈ ℝ
12 11 adantlr ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F ∧ x ∈ ran ⁡ F ∖ 0 → vol ⁡ F -1 x ∈ ℝ
13 10 12 remulcld ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F ∧ x ∈ ran ⁡ F ∖ 0 → x ⁢ vol ⁡ F -1 x ∈ ℝ
14 eldifi ⊢ x ∈ ran ⁡ F ∖ 0 → x ∈ ran ⁡ F
15 0cn ⊢ 0 ∈ ℂ
16 fnconstg ⊢ 0 ∈ ℂ → ℂ × 0 Fn ℂ
17 15 16 ax-mp ⊢ ℂ × 0 Fn ℂ
18 df-0p ⊢ 0 𝑝 = ℂ × 0
19 18 fneq1i ⊢ 0 𝑝 Fn ℂ ↔ ℂ × 0 Fn ℂ
20 17 19 mpbir ⊢ 0 𝑝 Fn ℂ
21 20 a1i ⊢ F ∈ dom ⁡ ∫ 1 → 0 𝑝 Fn ℂ
22 6 ffnd ⊢ F ∈ dom ⁡ ∫ 1 → F Fn ℝ
23 cnex ⊢ ℂ ∈ V
24 23 a1i ⊢ F ∈ dom ⁡ ∫ 1 → ℂ ∈ V
25 reex ⊢ ℝ ∈ V
26 25 a1i ⊢ F ∈ dom ⁡ ∫ 1 → ℝ ∈ V
27 ax-resscn ⊢ ℝ ⊆ ℂ
28 sseqin2 ⊢ ℝ ⊆ ℂ ↔ ℂ ∩ ℝ = ℝ
29 27 28 mpbi ⊢ ℂ ∩ ℝ = ℝ
30 0pval ⊢ y ∈ ℂ → 0 𝑝 ⁡ y = 0
31 30 adantl ⊢ F ∈ dom ⁡ ∫ 1 ∧ y ∈ ℂ → 0 𝑝 ⁡ y = 0
32 eqidd ⊢ F ∈ dom ⁡ ∫ 1 ∧ y ∈ ℝ → F ⁡ y = F ⁡ y
33 21 22 24 26 29 31 32 ofrfval ⊢ F ∈ dom ⁡ ∫ 1 → 0 𝑝 ≤ f F ↔ ∀ y ∈ ℝ 0 ≤ F ⁡ y
34 33 biimpa ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F → ∀ y ∈ ℝ 0 ≤ F ⁡ y
35 22 adantr ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F → F Fn ℝ
36 breq2 ⊢ x = F ⁡ y → 0 ≤ x ↔ 0 ≤ F ⁡ y
37 36 ralrn ⊢ F Fn ℝ → ∀ x ∈ ran ⁡ F 0 ≤ x ↔ ∀ y ∈ ℝ 0 ≤ F ⁡ y
38 35 37 syl ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F → ∀ x ∈ ran ⁡ F 0 ≤ x ↔ ∀ y ∈ ℝ 0 ≤ F ⁡ y
39 34 38 mpbird ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F → ∀ x ∈ ran ⁡ F 0 ≤ x
40 39 r19.21bi ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F ∧ x ∈ ran ⁡ F → 0 ≤ x
41 14 40 sylan2 ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F ∧ x ∈ ran ⁡ F ∖ 0 → 0 ≤ x
42 i1fima ⊢ F ∈ dom ⁡ ∫ 1 → F -1 x ∈ dom ⁡ vol
43 42 ad2antrr ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F ∧ x ∈ ran ⁡ F ∖ 0 → F -1 x ∈ dom ⁡ vol
44 mblss ⊢ F -1 x ∈ dom ⁡ vol → F -1 x ⊆ ℝ
45 ovolge0 ⊢ F -1 x ⊆ ℝ → 0 ≤ vol * ⁡ F -1 x
46 44 45 syl ⊢ F -1 x ∈ dom ⁡ vol → 0 ≤ vol * ⁡ F -1 x
47 mblvol ⊢ F -1 x ∈ dom ⁡ vol → vol ⁡ F -1 x = vol * ⁡ F -1 x
48 46 47 breqtrrd ⊢ F -1 x ∈ dom ⁡ vol → 0 ≤ vol ⁡ F -1 x
49 43 48 syl ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F ∧ x ∈ ran ⁡ F ∖ 0 → 0 ≤ vol ⁡ F -1 x
50 10 12 41 49 mulge0d ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F ∧ x ∈ ran ⁡ F ∖ 0 → 0 ≤ x ⁢ vol ⁡ F -1 x
51 5 13 50 fsumge0 ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F → 0 ≤ ∑ x ∈ ran ⁡ F ∖ 0 x ⁢ vol ⁡ F -1 x
52 itg1val ⊢ F ∈ dom ⁡ ∫ 1 → ∫ 1 ⁡ F = ∑ x ∈ ran ⁡ F ∖ 0 x ⁢ vol ⁡ F -1 x
53 52 adantr ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F → ∫ 1 ⁡ F = ∑ x ∈ ran ⁡ F ∖ 0 x ⁢ vol ⁡ F -1 x
54 51 53 breqtrrd ⊢ F ∈ dom ⁡ ∫ 1 ∧ 0 𝑝 ≤ f F → 0 ≤ ∫ 1 ⁡ F