Theorem brae 32188

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 484	. . . . 5 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → 𝑚 = 𝑀)
2	1	dmeqd 5811	. . . . . . 7 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → dom 𝑚 = dom 𝑀)
3	2	unieqd 4858	. . . . . 6 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → ∪ dom 𝑚 = ∪ dom 𝑀)
4		simpl 482	. . . . . 6 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → 𝑎 = 𝐴)
5	3, 4	difeq12d 4062	. . . . 5 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → (∪ dom 𝑚 ∖ 𝑎) = (∪ dom 𝑀 ∖ 𝐴))
6	1, 5	fveq12d 6775	. . . 4 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → (𝑚‘(∪ dom 𝑚 ∖ 𝑎)) = (𝑀‘(∪ dom 𝑀 ∖ 𝐴)))
7	6	eqeq1d 2741	. . 3 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → ((𝑚‘(∪ dom 𝑚 ∖ 𝑎)) = 0 ↔ (𝑀‘(∪ dom 𝑀 ∖ 𝐴)) = 0))
8		df-ae 32186	. . 3 ⊢ a.e. = {⟨𝑎, 𝑚⟩ ∣ (𝑚‘(∪ dom 𝑚 ∖ 𝑎)) = 0}
9	7, 8	brabga 5448	. 2 ⊢ ((𝐴 ∈ dom 𝑀 ∧ 𝑀 ∈ ∪ ran measures) → (𝐴a.e.𝑀 ↔ (𝑀‘(∪ dom 𝑀 ∖ 𝐴)) = 0))
10	9	ancoms 458	1 ⊢ ((𝑀 ∈ ∪ ran measures ∧ 𝐴 ∈ dom 𝑀) → (𝐴a.e.𝑀 ↔ (𝑀‘(∪ dom 𝑀 ∖ 𝐴)) = 0))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1541   ∈ wcel 2109   ∖ cdif 3888  ∪ cuni 4844   class class class wbr 5078  dom cdm 5588  ran crn 5589  ‘cfv 6430  0cc0 10855  measurescmeas 32142  a.e.cae 32184

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1801  ax-4 1815  ax-5 1916  ax-6 1974  ax-7 2014  ax-8 2111  ax-9 2119  ax-ext 2710  ax-sep 5226  ax-nul 5233  ax-pr 5355

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1544  df-fal 1554  df-ex 1786  df-sb 2071  df-clab 2717  df-cleq 2731  df-clel 2817  df-rab 3074  df-v 3432  df-dif 3894  df-un 3896  df-in 3898  df-ss 3908  df-nul 4262  df-if 4465  df-sn 4567  df-pr 4569  df-op 4573  df-uni 4845  df-br 5079  df-opab 5141  df-dm 5598  df-iota 6388  df-fv 6438  df-ae 32186

This theorem is referenced by: (None)