Theorem brae 34538

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

Assertion

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

Proof of Theorem brae

Dummy variables 𝑚 𝑎 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		simpr 488	. . . . 5 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → 𝑚 = 𝑀)
2	1	dmeqd 5881	. . . . . . 7 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → dom 𝑚 = dom 𝑀)
3	2	unieqd 4878	. . . . . 6 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → ∪ dom 𝑚 = ∪ dom 𝑀)
4		simpl 486	. . . . . 6 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → 𝑎 = 𝐴)
5	3, 4	difeq12d 4081	. . . . 5 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → (∪ dom 𝑚 ∖ 𝑎) = (∪ dom 𝑀 ∖ 𝐴))
6	1, 5	fveq12d 6874	. . . 4 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → (𝑚‘(∪ dom 𝑚 ∖ 𝑎)) = (𝑀‘(∪ dom 𝑀 ∖ 𝐴)))
7	6	eqeq1d 2764	. . 3 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → ((𝑚‘(∪ dom 𝑚 ∖ 𝑎)) = 0 ↔ (𝑀‘(∪ dom 𝑀 ∖ 𝐴)) = 0))
8		df-ae 34536	. . 3 ⊢ a.e. = {⟨𝑎, 𝑚⟩ ∣ (𝑚‘(∪ dom 𝑚 ∖ 𝑎)) = 0}
9	7, 8	brabga 5504	. 2 ⊢ ((𝐴 ∈ dom 𝑀 ∧ 𝑀 ∈ ∪ ran measures) → (𝐴a.e.𝑀 ↔ (𝑀‘(∪ dom 𝑀 ∖ 𝐴)) = 0))
10	9	ancoms 462	1 ⊢ ((𝑀 ∈ ∪ ran measures ∧ 𝐴 ∈ dom 𝑀) → (𝐴a.e.𝑀 ↔ (𝑀‘(∪ dom 𝑀 ∖ 𝐴)) = 0))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 399   = wceq 1560   ∈ wcel 2142   ∖ cdif 3901  ∪ cuni 4865   class class class wbr 5100  dom cdm 5647  ran crn 5648  ‘cfv 6521  0cc0 11073  measurescmeas 34492  a.e.cae 34534

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-ext 2734  ax-sep 5246  ax-pr 5390

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1100  df-tru 1563  df-fal 1573  df-ex 1800  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3456  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4481  df-sn 4583  df-pr 4585  df-op 4589  df-uni 4866  df-br 5101  df-opab 5163  df-dm 5657  df-iota 6477  df-fv 6529  df-ae 34536

This theorem is referenced by: (None)