Theorem brae 34204

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 5859	. . . . . . 7 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → dom 𝑚 = dom 𝑀)
3	2	unieqd 4880	. . . . . 6 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → ∪ dom 𝑚 = ∪ dom 𝑀)
4		simpl 482	. . . . . 6 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → 𝑎 = 𝐴)
5	3, 4	difeq12d 4086	. . . . 5 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → (∪ dom 𝑚 ∖ 𝑎) = (∪ dom 𝑀 ∖ 𝐴))
6	1, 5	fveq12d 6847	. . . 4 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → (𝑚‘(∪ dom 𝑚 ∖ 𝑎)) = (𝑀‘(∪ dom 𝑀 ∖ 𝐴)))
7	6	eqeq1d 2731	. . 3 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → ((𝑚‘(∪ dom 𝑚 ∖ 𝑎)) = 0 ↔ (𝑀‘(∪ dom 𝑀 ∖ 𝐴)) = 0))
8		df-ae 34202	. . 3 ⊢ a.e. = {⟨𝑎, 𝑚⟩ ∣ (𝑚‘(∪ dom 𝑚 ∖ 𝑎)) = 0}
9	7, 8	brabga 5489	. 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 206   ∧ wa 395   = wceq 1540   ∈ wcel 2109   ∖ cdif 3908  ∪ cuni 4867   class class class wbr 5102  dom cdm 5631  ran crn 5632  ‘cfv 6499  0cc0 11044  measurescmeas 34158  a.e.cae 34200

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701  ax-sep 5246  ax-nul 5256  ax-pr 5382

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-rab 3403  df-v 3446  df-dif 3914  df-un 3916  df-ss 3928  df-nul 4293  df-if 4485  df-sn 4586  df-pr 4588  df-op 4592  df-uni 4868  df-br 5103  df-opab 5165  df-dm 5641  df-iota 6452  df-fv 6507  df-ae 34202

This theorem is referenced by: (None)