Metamath Proof Explorer


Theorem brae

Description: 'almost everywhere' relation for a measure and a measurable set A . (Contributed by Thierry Arnoux, 20-Oct-2017)

Ref Expression
Assertion brae ( ( 𝑀 ∈ ∪ ran measures ∧ 𝐴 ∈ dom 𝑀 ) → ( 𝐴 a.e. 𝑀 ↔ ( 𝑀 ‘ ( ∪ dom 𝑀 ∖ 𝐴 ) ) = 0 ) )

Proof

Step Hyp Ref Expression
1 simpr ⊢ ( ( 𝑎 = 𝐴 ∧ 𝑚 = 𝑀 ) → 𝑚 = 𝑀 )
2 1 dmeqd ⊢ ( ( 𝑎 = 𝐴 ∧ 𝑚 = 𝑀 ) → dom 𝑚 = dom 𝑀 )
3 2 unieqd ⊢ ( ( 𝑎 = 𝐴 ∧ 𝑚 = 𝑀 ) → ∪ dom 𝑚 = ∪ dom 𝑀 )
4 simpl ⊢ ( ( 𝑎 = 𝐴 ∧ 𝑚 = 𝑀 ) → 𝑎 = 𝐴 )
5 3 4 difeq12d ⊢ ( ( 𝑎 = 𝐴 ∧ 𝑚 = 𝑀 ) → ( ∪ dom 𝑚 ∖ 𝑎 ) = ( ∪ dom 𝑀 ∖ 𝐴 ) )
6 1 5 fveq12d ⊢ ( ( 𝑎 = 𝐴 ∧ 𝑚 = 𝑀 ) → ( 𝑚 ‘ ( ∪ dom 𝑚 ∖ 𝑎 ) ) = ( 𝑀 ‘ ( ∪ dom 𝑀 ∖ 𝐴 ) ) )
7 6 eqeq1d ⊢ ( ( 𝑎 = 𝐴 ∧ 𝑚 = 𝑀 ) → ( ( 𝑚 ‘ ( ∪ dom 𝑚 ∖ 𝑎 ) ) = 0 ↔ ( 𝑀 ‘ ( ∪ dom 𝑀 ∖ 𝐴 ) ) = 0 ) )
8 df-ae ⊢ a.e. = { ⟨ 𝑎 , 𝑚 ⟩ ∣ ( 𝑚 ‘ ( ∪ dom 𝑚 ∖ 𝑎 ) ) = 0 }
9 7 8 brabga ⊢ ( ( 𝐴 ∈ dom 𝑀 ∧ 𝑀 ∈ ∪ ran measures ) → ( 𝐴 a.e. 𝑀 ↔ ( 𝑀 ‘ ( ∪ dom 𝑀 ∖ 𝐴 ) ) = 0 ) )
10 9 ancoms ⊢ ( ( 𝑀 ∈ ∪ ran measures ∧ 𝐴 ∈ dom 𝑀 ) → ( 𝐴 a.e. 𝑀 ↔ ( 𝑀 ‘ ( ∪ dom 𝑀 ∖ 𝐴 ) ) = 0 ) )