let X be non empty set ; for S being SigmaField of X
for M being sigma_Measure of S
for f being PartFunc of X,ExtREAL
for A being Element of S st ex E being Element of S st
( E = dom f & f is E -measurable ) & ( for x being object st x in (dom f) \ A holds
f . x = 0 ) & f is nonnegative holds
Integral (M,f) = Integral (M,(f | A))
let S be SigmaField of X; for M being sigma_Measure of S
for f being PartFunc of X,ExtREAL
for A being Element of S st ex E being Element of S st
( E = dom f & f is E -measurable ) & ( for x being object st x in (dom f) \ A holds
f . x = 0 ) & f is nonnegative holds
Integral (M,f) = Integral (M,(f | A))
let M be sigma_Measure of S; for f being PartFunc of X,ExtREAL
for A being Element of S st ex E being Element of S st
( E = dom f & f is E -measurable ) & ( for x being object st x in (dom f) \ A holds
f . x = 0 ) & f is nonnegative holds
Integral (M,f) = Integral (M,(f | A))
let f be PartFunc of X,ExtREAL; for A being Element of S st ex E being Element of S st
( E = dom f & f is E -measurable ) & ( for x being object st x in (dom f) \ A holds
f . x = 0 ) & f is nonnegative holds
Integral (M,f) = Integral (M,(f | A))
let A be Element of S; ( ex E being Element of S st
( E = dom f & f is E -measurable ) & ( for x being object st x in (dom f) \ A holds
f . x = 0 ) & f is nonnegative implies Integral (M,f) = Integral (M,(f | A)) )
assume that
A1:
ex E being Element of S st
( E = dom f & f is E -measurable )
and
A2:
for x being object st x in (dom f) \ A holds
f . x = 0
and
A3:
f is nonnegative
; Integral (M,f) = Integral (M,(f | A))
consider E being Element of S such that
A4:
( E = dom f & f is E -measurable )
by A1;
reconsider B = E \ A as Element of S ;
A6:
Integral (M,(f | (A \/ B))) = (Integral (M,(f | A))) + (Integral (M,(f | B)))
by A1, A3, MESFUNC5:91, XBOOLE_1:79;
A \/ B = (dom f) \/ A
by A4, XBOOLE_1:39;
then A7:
f | (A \/ B) = f
by RELAT_1:68, XBOOLE_1:7;
A8:
B c= dom f
by A4, XBOOLE_1:36;
dom (f | B) = (dom f) /\ B
by RELAT_1:61;
then A9:
dom (f | B) = B
by A4, XBOOLE_1:28, XBOOLE_1:36;
then Integral (M,(f | B)) =
0 * (M . (dom (f | B)))
by A9, Th27
.=
0
;
hence
Integral (M,f) = Integral (M,(f | A))
by A6, A7, XXREAL_3:4; verum