Metamath Proof Explorer


Theorem ismea

Description: Express the predicate " M is a measure." Definition 112A of Fremlin1 p. 14. (Contributed by Glauco Siliprandi, 17-Aug-2020)

Ref Expression
Assertion ismea M Meas M : dom M 0 +∞ dom M SAlg M = 0 x 𝒫 dom M x ω Disj y x y M x = sum^ M x

Proof

Step Hyp Ref Expression
1 id M Meas M Meas
2 fex M : dom M 0 +∞ dom M SAlg M V
3 id z = M z = M
4 dmeq z = M dom z = dom M
5 3 4 feq12d z = M z : dom z 0 +∞ M : dom M 0 +∞
6 4 eleq1d z = M dom z SAlg dom M SAlg
7 5 6 anbi12d z = M z : dom z 0 +∞ dom z SAlg M : dom M 0 +∞ dom M SAlg
8 fveq1 z = M z = M
9 8 eqeq1d z = M z = 0 M = 0
10 7 9 anbi12d z = M z : dom z 0 +∞ dom z SAlg z = 0 M : dom M 0 +∞ dom M SAlg M = 0
11 4 pweqd z = M 𝒫 dom z = 𝒫 dom M
12 fveq1 z = M z x = M x
13 reseq1 z = M z x = M x
14 13 fveq2d z = M sum^ z x = sum^ M x
15 12 14 eqeq12d z = M z x = sum^ z x M x = sum^ M x
16 15 imbi2d z = M x ω Disj y x y z x = sum^ z x x ω Disj y x y M x = sum^ M x
17 11 16 raleqbidv z = M x 𝒫 dom z x ω Disj y x y z x = sum^ z x x 𝒫 dom M x ω Disj y x y M x = sum^ M x
18 10 17 anbi12d z = M z : dom z 0 +∞ dom z SAlg z = 0 x 𝒫 dom z x ω Disj y x y z x = sum^ z x M : dom M 0 +∞ dom M SAlg M = 0 x 𝒫 dom M x ω Disj y x y M x = sum^ M x
19 df-mea Meas = z | z : dom z 0 +∞ dom z SAlg z = 0 x 𝒫 dom z x ω Disj y x y z x = sum^ z x
20 18 19 elab2g M V M Meas M : dom M 0 +∞ dom M SAlg M = 0 x 𝒫 dom M x ω Disj y x y M x = sum^ M x
21 2 20 syl M : dom M 0 +∞ dom M SAlg M Meas M : dom M 0 +∞ dom M SAlg M = 0 x 𝒫 dom M x ω Disj y x y M x = sum^ M x
22 21 ad2antrr M : dom M 0 +∞ dom M SAlg M = 0 x 𝒫 dom M x ω Disj y x y M x = sum^ M x M Meas M : dom M 0 +∞ dom M SAlg M = 0 x 𝒫 dom M x ω Disj y x y M x = sum^ M x
23 22 ibir M : dom M 0 +∞ dom M SAlg M = 0 x 𝒫 dom M x ω Disj y x y M x = sum^ M x M Meas
24 18 19 elab2g M Meas M Meas M : dom M 0 +∞ dom M SAlg M = 0 x 𝒫 dom M x ω Disj y x y M x = sum^ M x
25 1 23 24 pm5.21nii M Meas M : dom M 0 +∞ dom M SAlg M = 0 x 𝒫 dom M x ω Disj y x y M x = sum^ M x