Metamath Proof Explorer


Theorem smfdivdmmbl

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 needed only for the function at the denominator). (Contributed by Glauco Siliprandi, 5-Jan-2025)

Ref Expression
Hypotheses smfdivdmmbl.1 x φ
smfdivdmmbl.2 _ x B
smfdivdmmbl.3 φ S SAlg
smfdivdmmbl.4 φ A S
smfdivdmmbl.5 φ B S
smfdivdmmbl.6 φ x B D W
smfdivdmmbl.7 φ x B D SMblFn S
smfdivdmmbl.8 E = x B | D 0
Assertion smfdivdmmbl φ A E S

Proof

Step Hyp Ref Expression
1 smfdivdmmbl.1 x φ
2 smfdivdmmbl.2 _ x B
3 smfdivdmmbl.3 φ S SAlg
4 smfdivdmmbl.4 φ A S
5 smfdivdmmbl.5 φ B S
6 smfdivdmmbl.6 φ x B D W
7 smfdivdmmbl.7 φ x B D SMblFn S
8 smfdivdmmbl.8 E = x B | D 0
9 nfcv _ x
10 1 2 3 6 7 smffmptf φ x B D : B
11 2 9 10 fvmptelcdmf φ x B D
12 0red φ 0
13 1 2 3 5 11 7 12 8 smfdmmblpimne φ E S
14 3 4 13 salincld φ A E S