let X be non empty set ; :: thesis: for S being SigmaField of X
for M being sigma_Measure of S
for f being PartFunc of X,COMPLEX
for A being Element of S st ex E being Element of S st
( E = dom f & f is E -measurable ) & M . A = 0 holds
( f | A is_integrable_on M & Integral (M,(f | A)) = 0 )

let S be SigmaField of X; :: thesis: for M being sigma_Measure of S
for f being PartFunc of X,COMPLEX
for A being Element of S st ex E being Element of S st
( E = dom f & f is E -measurable ) & M . A = 0 holds
( f | A is_integrable_on M & Integral (M,(f | A)) = 0 )

let M be sigma_Measure of S; :: thesis: for f being PartFunc of X,COMPLEX
for A being Element of S st ex E being Element of S st
( E = dom f & f is E -measurable ) & M . A = 0 holds
( f | A is_integrable_on M & Integral (M,(f | A)) = 0 )

let f be PartFunc of X,COMPLEX; :: thesis: for A being Element of S st ex E being Element of S st
( E = dom f & f is E -measurable ) & M . A = 0 holds
( f | A is_integrable_on M & Integral (M,(f | A)) = 0 )

let A be Element of S; :: thesis: ( ex E being Element of S st
( E = dom f & f is E -measurable ) & M . A = 0 implies ( f | A is_integrable_on M & Integral (M,(f | A)) = 0 ) )

set g = f | A;
assume that
A1: ex E being Element of S st
( E = dom f & f is E -measurable ) and
A2: M . A = 0 ; :: thesis: ( f | A is_integrable_on M & Integral (M,(f | A)) = 0 )
consider E being Element of S such that
A3: E = dom f and
A4: f is E -measurable by A1;
A5: dom (Im f) = dom f by COMSEQ_3:def 4;
A6: Im f is E -measurable by A4;
then A7: Integral (M,((Im f) | A)) = 0 by ;
(Im f) | A is_integrable_on M by A2, A3, A6, A5, Th20;
then A8: Im (f | A) is_integrable_on M by Th7;
A9: dom (Re f) = dom f by COMSEQ_3:def 3;
A10: Im (f | A) = (Im f) | A by Th7;
A11: Re f is E -measurable by A4;
A12: Re (f | A) = (Re f) | A by Th7;
then reconsider R = Integral (M,(Re (f | A))), I = Integral (M,(Im (f | A))) as Real by A2, A3, A11, A6, A9, A5, A10, MESFUNC6:88;
(Re f) | A is_integrable_on M by A2, A3, A11, A9, Th20;
then Re (f | A) is_integrable_on M by Th7;
hence f | A is_integrable_on M by A8; :: thesis: Integral (M,(f | A)) = 0
hence Integral (M,(f | A)) = R + (I * <i>) by Def3
.= 0 by A2, A3, A11, A9, A7, A12, A10, MESFUNC6:88 ;
:: thesis: verum