Theorem brae 34398

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 5854	. . . . . . 7 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → dom 𝑚 = dom 𝑀)
3	2	unieqd 4876	. . . . . 6 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → ∪ dom 𝑚 = ∪ dom 𝑀)
4		simpl 482	. . . . . 6 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → 𝑎 = 𝐴)
5	3, 4	difeq12d 4079	. . . . 5 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → (∪ dom 𝑚 ∖ 𝑎) = (∪ dom 𝑀 ∖ 𝐴))
6	1, 5	fveq12d 6841	. . . 4 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → (𝑚‘(∪ dom 𝑚 ∖ 𝑎)) = (𝑀‘(∪ dom 𝑀 ∖ 𝐴)))
7	6	eqeq1d 2738	. . 3 ⊢ ((𝑎 = 𝐴 ∧ 𝑚 = 𝑀) → ((𝑚‘(∪ dom 𝑚 ∖ 𝑎)) = 0 ↔ (𝑀‘(∪ dom 𝑀 ∖ 𝐴)) = 0))
8		df-ae 34396	. . 3 ⊢ a.e. = {⟨𝑎, 𝑚⟩ ∣ (𝑚‘(∪ dom 𝑚 ∖ 𝑎)) = 0}
9	7, 8	brabga 5482	. 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 1541   ∈ wcel 2113   ∖ cdif 3898  ∪ cuni 4863   class class class wbr 5098  dom cdm 5624  ran crn 5625  ‘cfv 6492  0cc0 11026  measurescmeas 34352  a.e.cae 34394

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2708  ax-sep 5241  ax-nul 5251  ax-pr 5377

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2715  df-cleq 2728  df-clel 2811  df-rab 3400  df-v 3442  df-dif 3904  df-un 3906  df-ss 3918  df-nul 4286  df-if 4480  df-sn 4581  df-pr 4583  df-op 4587  df-uni 4864  df-br 5099  df-opab 5161  df-dm 5634  df-iota 6448  df-fv 6500  df-ae 34396

This theorem is referenced by: (None)