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_measurable_on E ) & 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_measurable_on E ) & 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_measurable_on E ) & 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_measurable_on E ) & 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_measurable_on E ) & M . A = 0 implies ( f | A is_integrable_on M & Integral M,(f | A) = 0 ) )
set g = f | A;
assume A1: ( ex E being Element of S st
( E = dom f & f is_measurable_on E ) & M . A = 0 ) ; :: thesis: ( f | A is_integrable_on M & Integral M,(f | A) = 0 )
then consider E being Element of S such that
A2: ( E = dom f & f is_measurable_on E ) ;
A3: ( Re f is_measurable_on E & Im f is_measurable_on E ) by A2, Def3;
A4: ( dom (Re f) = dom f & dom (Im f) = dom f ) by Def1, Def2;
then A5: ( Integral M,((Re f) | A) = 0 & Integral M,((Im f) | A) = 0 ) by A1, A2, A3, MESFUNC6:88;
A6: ( Re (f | A) = (Re f) | A & Im (f | A) = (Im f) | A ) by COM91;
then reconsider R = Integral M,(Re (f | A)), I = Integral M,(Im (f | A)) as Real by A4, A2, A1, A3, MESFUNC6:88;
( (Re f) | A is_integrable_on M & (Im f) | A is_integrable_on M ) by A1, A3, A4, A2, Th88a;
then ( Re (f | A) is_integrable_on M & Im (f | A) is_integrable_on M ) by COM91;
hence f | A is_integrable_on M by Def4; :: thesis: Integral M,(f | A) = 0
hence Integral M,(f | A) = R + (I * <i> ) by Def5
.= 0 by A5, A6 ;
:: thesis: verum