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 xφ
smfdivdmmbl.2 _xB
smfdivdmmbl.3 φSSAlg
smfdivdmmbl.4 φAS
smfdivdmmbl.5 φBS
smfdivdmmbl.6 φxBDW
smfdivdmmbl.7 φxBDSMblFnS
smfdivdmmbl.8 E=xB|D0
Assertion smfdivdmmbl φAES

Proof

Step Hyp Ref Expression
1 smfdivdmmbl.1 xφ
2 smfdivdmmbl.2 _xB
3 smfdivdmmbl.3 φSSAlg
4 smfdivdmmbl.4 φAS
5 smfdivdmmbl.5 φBS
6 smfdivdmmbl.6 φxBDW
7 smfdivdmmbl.7 φxBDSMblFnS
8 smfdivdmmbl.8 E=xB|D0
9 nfcv _x
10 1 2 3 6 7 smffmptf φxBD:B
11 2 9 10 fvmptelcdmf φxBD
12 0red φ0
13 1 2 3 5 11 7 12 8 smfdmmblpimne φES
14 3 4 13 salincld φAES