Metamath Proof Explorer


Theorem smfdivdmmbl2

Description: If a functions and a sigma-measurable function have domains in the sigma-algebra, the domain of the division of the two functions is in the sigma-algebra. This is the third statement of Proposition 121H of Fremlin1 p. 39 . Note: While the theorem in the book assumes both functions are sigma-measurable, this assumption is unnecessary for the part concerning their division, for the function at the numerator. It is required only for the function at the denominator. (Contributed by Glauco Siliprandi, 5-Jan-2025)

Ref Expression
Hypotheses smfdivdmmbl2.1 ⊢ Ⅎ x φ
smfdivdmmbl2.2 ⊢ Ⅎ _ x F
smfdivdmmbl2.3 ⊢ Ⅎ _ x G
smfdivdmmbl2.4 ⊢ φ → S ∈ SAlg
smfdivdmmbl2.5 ⊢ φ → F : A ⟶ V
smfdivdmmbl2.6 ⊢ φ → G ∈ SMblFn ⁡ S
smfdivdmmbl2.7 ⊢ φ → A ∈ S
smfdivdmmbl2.8 ⊢ φ → dom ⁡ G ∈ S
smfdivdmmbl2.9 ⊢ D = x ∈ dom ⁡ G | G ⁡ x ≠ 0
smfdivdmmbl2.10 ⊢ H = x ∈ dom ⁡ F ∩ D ⟼ F ⁡ x G ⁡ x
Assertion smfdivdmmbl2 ⊢ φ → dom ⁡ H ∈ S

Proof

Step Hyp Ref Expression
1 smfdivdmmbl2.1 ⊢ Ⅎ x φ
2 smfdivdmmbl2.2 ⊢ Ⅎ _ x F
3 smfdivdmmbl2.3 ⊢ Ⅎ _ x G
4 smfdivdmmbl2.4 ⊢ φ → S ∈ SAlg
5 smfdivdmmbl2.5 ⊢ φ → F : A ⟶ V
6 smfdivdmmbl2.6 ⊢ φ → G ∈ SMblFn ⁡ S
7 smfdivdmmbl2.7 ⊢ φ → A ∈ S
8 smfdivdmmbl2.8 ⊢ φ → dom ⁡ G ∈ S
9 smfdivdmmbl2.9 ⊢ D = x ∈ dom ⁡ G | G ⁡ x ≠ 0
10 smfdivdmmbl2.10 ⊢ H = x ∈ dom ⁡ F ∩ D ⟼ F ⁡ x G ⁡ x
11 2 nfdm ⊢ Ⅎ _ x dom ⁡ F
12 nfrab1 ⊢ Ⅎ _ x x ∈ dom ⁡ G | G ⁡ x ≠ 0
13 9 12 nfcxfr ⊢ Ⅎ _ x D
14 11 13 nfin ⊢ Ⅎ _ x dom ⁡ F ∩ D
15 ovex ⊢ F ⁡ x G ⁡ x ∈ V
16 14 15 10 dmmptif ⊢ dom ⁡ H = dom ⁡ F ∩ D
17 5 fdmd ⊢ φ → dom ⁡ F = A
18 17 7 eqeltrd ⊢ φ → dom ⁡ F ∈ S
19 4 8 salrestss ⊢ φ → S ↾ 𝑡 dom ⁡ G ⊆ S
20 eqid ⊢ dom ⁡ G = dom ⁡ G
21 1 3 4 6 20 smfpimne2 ⊢ φ → x ∈ dom ⁡ G | G ⁡ x ≠ 0 ∈ S ↾ 𝑡 dom ⁡ G
22 9 21 eqeltrid ⊢ φ → D ∈ S ↾ 𝑡 dom ⁡ G
23 19 22 sseldd ⊢ φ → D ∈ S
24 4 18 23 salincld ⊢ φ → dom ⁡ F ∩ D ∈ S
25 16 24 eqeltrid ⊢ φ → dom ⁡ H ∈ S