let X be non empty set ; :: thesis: for S being SigmaField of X
for M being sigma_Measure of S
for E being Element of S
for F being Functional_Sequence of X,ExtREAL
for f being PartFunc of X,ExtREAL st E c= dom f & f is_measurable_on E & E = {} & ( for n being Nat holds F . n is_simple_func_in S ) holds
ex I being ExtREAL_sequence st
( ( for n being Nat holds I . n = Integral M,((F . n) | E) ) & I is summable & Integral M,(f | E) = Sum I )

let S be SigmaField of X; :: thesis: for M being sigma_Measure of S
for E being Element of S
for F being Functional_Sequence of X,ExtREAL
for f being PartFunc of X,ExtREAL st E c= dom f & f is_measurable_on E & E = {} & ( for n being Nat holds F . n is_simple_func_in S ) holds
ex I being ExtREAL_sequence st
( ( for n being Nat holds I . n = Integral M,((F . n) | E) ) & I is summable & Integral M,(f | E) = Sum I )

let M be sigma_Measure of S; :: thesis: for E being Element of S
for F being Functional_Sequence of X,ExtREAL
for f being PartFunc of X,ExtREAL st E c= dom f & f is_measurable_on E & E = {} & ( for n being Nat holds F . n is_simple_func_in S ) holds
ex I being ExtREAL_sequence st
( ( for n being Nat holds I . n = Integral M,((F . n) | E) ) & I is summable & Integral M,(f | E) = Sum I )

let E be Element of S; :: thesis: for F being Functional_Sequence of X,ExtREAL
for f being PartFunc of X,ExtREAL st E c= dom f & f is_measurable_on E & E = {} & ( for n being Nat holds F . n is_simple_func_in S ) holds
ex I being ExtREAL_sequence st
( ( for n being Nat holds I . n = Integral M,((F . n) | E) ) & I is summable & Integral M,(f | E) = Sum I )

let F be Functional_Sequence of X,ExtREAL ; :: thesis: for f being PartFunc of X,ExtREAL st E c= dom f & f is_measurable_on E & E = {} & ( for n being Nat holds F . n is_simple_func_in S ) holds
ex I being ExtREAL_sequence st
( ( for n being Nat holds I . n = Integral M,((F . n) | E) ) & I is summable & Integral M,(f | E) = Sum I )

let f be PartFunc of X,ExtREAL ; :: thesis: ( E c= dom f & f is_measurable_on E & E = {} & ( for n being Nat holds F . n is_simple_func_in S ) implies ex I being ExtREAL_sequence st
( ( for n being Nat holds I . n = Integral M,((F . n) | E) ) & I is summable & Integral M,(f | E) = Sum I ) )

assume that
A0: E c= dom f and
B2: f is_measurable_on E and
A1: E = {} and
A2: for n being Nat holds F . n is_simple_func_in S ; :: thesis: ex I being ExtREAL_sequence st
( ( for n being Nat holds I . n = Integral M,((F . n) | E) ) & I is summable & Integral M,(f | E) = Sum I )

take I = NAT --> 0. ; :: thesis: ( ( for n being Nat holds I . n = Integral M,((F . n) | E) ) & I is summable & Integral M,(f | E) = Sum I )
A5: M . E = 0 by A1, VALUED_0:def 19;
thus for n being Nat holds I . n = Integral M,((F . n) | E) :: thesis: ( I is summable & Integral M,(f | E) = Sum I )
proof
let n be Nat; :: thesis: I . n = Integral M,((F . n) | E)
reconsider m = n as Element of NAT by ORDINAL1:def 13;
reconsider D = dom (F . m) as Element of S by A2, MESFUNC5:43;
F . m is_measurable_on D by A2, MESFUNC2:37;
then Integral M,((F . m) | E) = 0 by A5, MESFUNC5:100;
hence I . n = Integral M,((F . n) | E) by FUNCOP_1:13; :: thesis: verum
end;
defpred S1[ Element of NAT ] means (Partial_Sums I) . $1 = 0 ;
(Partial_Sums I) . 0 = I . 0 by Def1;
then P1: S1[ 0 ] by FUNCOP_1:13;
P2: for k being Element of NAT st S1[k] holds
S1[k + 1]
proof
let k be Element of NAT ; :: thesis: ( S1[k] implies S1[k + 1] )
assume P21: S1[k] ; :: thesis: S1[k + 1]
P22: I . (k + 1) = 0 by FUNCOP_1:13;
(Partial_Sums I) . (k + 1) = ((Partial_Sums I) . k) + (I . (k + 1)) by Def1;
then (Partial_Sums I) . (k + 1) = 0 + 0 by P21, P22;
hence S1[k + 1] ; :: thesis: verum
end;
P3: for k being Element of NAT holds S1[k] from NAT_1:sch 1(P1, P2);
for n being Nat holds (Partial_Sums I) . n = 0
proof end;
then A6: ( Partial_Sums I is convergent_to_finite_number & lim (Partial_Sums I) = 0 ) by MESFUNC5:58;
then Partial_Sums I is convergent by MESFUNC5:def 11;
hence I is summable by Def2; :: thesis: Integral M,(f | E) = Sum I
A7: E = dom (f | E) by A0, RELAT_1:91;
then E = (dom f) /\ E by RELAT_1:90;
then Integral M,((f | E) | E) = 0 by A7, A5, B2, MESFUNC5:48, MESFUNC5:100;
hence Integral M,(f | E) = Sum I by A6, FUNCT_1:82; :: thesis: verum