let X be non empty set ; :: thesis:
let S be SigmaField of X; :: thesis:
let M be sigma_Measure of S; :: thesis:
let f be PartFunc of X,ExtREAL; :: thesis:
assume that
A1: f is_simple_func_in S and
A2: f is nonnegative ; :: thesis:
set A = dom f;
set g = f | ((dom f) /\ (eq_dom (f,0)));
for x being object st x in eq_dom (f,0) holds
x in dom f by MESFUNC1:def 15;
then eq_dom (f,0) c= dom f ;
then A3: f | ((dom f) /\ (eq_dom (f,0))) = f | (eq_dom (f,0)) by XBOOLE_1:28;
A4: ex G being Finite_Sep_Sequence of S st
( dom (f | ((dom f) /\ (eq_dom (f,0)))) = union (rng G) & ( for n being Nat
for x, y being Element of X st n in dom G & x in G . n & y in G . n holds
(f | ((dom f) /\ (eq_dom (f,0)))) . x = (f | ((dom f) /\ (eq_dom (f,0)))) . y ) )

consider F being Finite_Sep_Sequence of S such that
A5: dom f = union (rng F) and
A6: for n being Nat
for x, y being Element of X st n in dom F & x in F . n & y in F . n holds
f . x = f . y by A1, MESFUNC2:def 4;
deffunc H₁( Nat) -> Element of bool X = (F . $1) /\ ((dom f) /\ (eq_dom (f,0)));
reconsider A = dom f as Element of S by A5, MESFUNC2:31;
consider G being FinSequence such that
A7: ( len G = len F & ( for n being Nat st n in dom G holds
G . n = H₁(n) ) ) from FINSEQ_1:sch 2();
f is A -measurable by A1, MESFUNC2:34;
then A /\ (less_dom (f,0)) in S by MESFUNC1:def 16;
then A \ (A /\ (less_dom (f,0))) in S by PROB_1:6;
then reconsider A1 = A /\ (great_eq_dom (f,0.)) as Element of S by MESFUNC1:14;
f is A1 -measurable by A1, MESFUNC2:34;
then (A /\ (great_eq_dom (f,0))) /\ (less_eq_dom (f,0)) in S by MESFUNC1:28;
then reconsider A2 = A /\ (eq_dom (f,0)) as Element of S by MESFUNC1:18;
A8: dom F = Seg (len F) by FINSEQ_1:def 3;
dom G = Seg (len F) by A7, FINSEQ_1:def 3;
then A9: for i being Nat st i in dom F holds
G . i = (F . i) /\ A2 by A7, A8;
dom G = Seg (len F) by A7, FINSEQ_1:def 3;
then A10: dom G = dom F by FINSEQ_1:def 3;
then reconsider G = G as Finite_Sep_Sequence of S by A9, Th35;
take G ; :: thesis:
for i being Nat st i in dom G holds
G . i = A2 /\ (F . i) by A7;
then A11: union (rng G) = A2 /\ (dom f) by A5, A10, MESFUNC3:6
.= dom (f | ((dom f) /\ (eq_dom (f,0)))) by RELAT_1:61 ;
for i being Nat
for x, y being Element of X st i in dom G & x in G . i & y in G . i holds
(f | ((dom f) /\ (eq_dom (f,0)))) . x = (f | ((dom f) /\ (eq_dom (f,0)))) . y

let i be Nat; :: thesis:
let x, y be Element of X; :: thesis:
assume that
A12: i in dom G and
A13: x in G . i and
A14: y in G . i ; :: thesis:
A15: G . i = (F . i) /\ A2 by A7, A12;
then A16: y in F . i by A14, XBOOLE_0:def 4;
A17: G . i in rng G by A12, FUNCT_1:3;
then x in dom (f | ((dom f) /\ (eq_dom (f,0)))) by A11, A13, TARSKI:def 4;
then A18: (f | ((dom f) /\ (eq_dom (f,0)))) . x = f . x by FUNCT_1:47;
y in dom (f | ((dom f) /\ (eq_dom (f,0)))) by A11, A14, A17, TARSKI:def 4;
then A19: (f | ((dom f) /\ (eq_dom (f,0)))) . y = f . y by FUNCT_1:47;
x in F . i by A13, A15, XBOOLE_0:def 4;
hence (f | ((dom f) /\ (eq_dom (f,0)))) . x = (f | ((dom f) /\ (eq_dom (f,0)))) . y by A6, A10, A12, A16, A18, A19; :: thesis:

end;

hence ( dom (f | ((dom f) /\ (eq_dom (f,0)))) = union (rng G) & ( for n being Nat
for x, y being Element of X st n in dom G & x in G . n & y in G . n holds
(f | ((dom f) /\ (eq_dom (f,0)))) . x = (f | ((dom f) /\ (eq_dom (f,0)))) . y ) ) by A11; :: thesis:

end;

for x being object st x in dom (f | ((dom f) /\ (eq_dom (f,0)))) holds
0 <= (f | ((dom f) /\ (eq_dom (f,0)))) . x

proof

let x be object ; :: thesis:
assume A21: x in dom (f | ((dom f) /\ (eq_dom (f,0)))) ; :: thesis:
0 <= f . x by A2, SUPINF_2:51;
hence 0 <= (f | ((dom f) /\ (eq_dom (f,0)))) . x by A21, FUNCT_1:47; :: thesis:

end;

then a2: f | ((dom f) /\ (eq_dom (f,0))) is nonnegative by SUPINF_2:52;
f is real-valued by A1, MESFUNC2:def 4;
then A22: f | ((dom f) /\ (eq_dom (f,0))) is_simple_func_in S by A4, MESFUNC2:def 4;

now :: thesis:

consider F being Finite_Sep_Sequence of S, a, x being FinSequence of ExtREAL such that
A23: F,a are_Re-presentation_of f | ((dom f) /\ (eq_dom (f,0))) and
a . 1 = 0 and
for n being Nat st 2 <= n & n in dom a holds
( 0 < a . n & a . n < +infty ) and
A24: dom x = dom F and
A25: for n being Nat st n in dom x holds
x . n = (a . n) * ((M * F) . n) and
A26: integral (M,(f | ((dom f) /\ (eq_dom (f,0))))) = Sum x by a2, A22, MESFUNC3:def 2;
A27: for x being set st x in dom (f | ((dom f) /\ (eq_dom (f,0)))) holds
(f | ((dom f) /\ (eq_dom (f,0)))) . x = 0

proof

let x be set ; :: thesis:
assume A28: x in dom (f | ((dom f) /\ (eq_dom (f,0)))) ; :: thesis:
then x in (dom f) /\ ((dom f) /\ (eq_dom (f,0))) by RELAT_1:61;
then x in (dom f) /\ (eq_dom (f,0)) by XBOOLE_0:def 4;
then x in eq_dom (f,0) by XBOOLE_0:def 4;
then 0 = f . x by MESFUNC1:def 15;
hence (f | ((dom f) /\ (eq_dom (f,0)))) . x = 0 by A28, FUNCT_1:47; :: thesis:

end;

A29: for n being Nat holds
( not n in dom F or a . n = 0 or F . n = {} )

proof

let n be Nat; :: thesis:
assume A30: n in dom F ; :: thesis:

now :: thesis:

assume F . n <> {} ; :: thesis:
then consider x being object such that
A31: x in F . n by XBOOLE_0:def 1;
F . n in rng F by A30, FUNCT_1:3;
then x in union (rng F) by A31, TARSKI:def 4;
then x in dom (f | ((dom f) /\ (eq_dom (f,0)))) by A23, MESFUNC3:def 1;
then (f | ((dom f) /\ (eq_dom (f,0)))) . x = 0 by A27;
hence ( a . n = 0 or F . n = {} ) by A23, A30, A31, MESFUNC3:def 1; :: thesis:

end;

hence ( a . n = 0 or F . n = {} ) ; :: thesis:

end;

A32: for n being Nat st n in dom x holds
x . n = 0

proof

let n be Nat; :: thesis:
assume A33: n in dom x ; :: thesis:

per cases by A24, A29, A33;

suppose a . n = 0 ; :: thesis:

then (a . n) * ((M * F) . n) = 0 ;
hence x . n = 0 by A25, A33; :: thesis:

end;

suppose F . n = {} ; :: thesis:

then M . (F . n) = 0 by VALUED_0:def 19;
then (M * F) . n = 0 by A24, A33, FUNCT_1:13;
then (a . n) * ((M * F) . n) = 0 ;
hence x . n = 0 by A25, A33; :: thesis:

end;

A34: Sum x = 0

proof

consider sumx being sequence of ExtREAL such that
A35: Sum x = sumx . (len x) and
A36: sumx . 0 = 0 and
A37: for i being Nat st i < len x holds
sumx . (i + 1) = (sumx . i) + (x . (i + 1)) by EXTREAL1:def 2;

now :: thesis:

defpred S₁[ Nat] means ( $1 <= len x implies sumx . $1 = 0 );
assume x <> {} ; :: thesis:
A38: for k being Nat st S₁[k] holds
S₁[k + 1]

proof

let k be Nat; :: thesis:
assume A39: S₁[k] ; :: thesis:
assume A40: k + 1 <= len x ; :: thesis:
reconsider k = k as Element of NAT by ORDINAL1:def 12;
1 <= k + 1 by NAT_1:11;
then k + 1 in Seg (len x) by A40;
then k + 1 in dom x by FINSEQ_1:def 3;
then A41: x . (k + 1) = 0 by A32;
k < len x by A40, NAT_1:13;
then sumx . (k + 1) = (sumx . k) + (x . (k + 1)) by A37;
hence sumx . (k + 1) = 0 by A39, A40, A41, NAT_1:13; :: thesis:

end;

A42: S₁[ 0 ] by A36;
for i being Nat holds S₁[i] from NAT_1:sch 2(A42, A38);
hence Sum x = 0 by A35; :: thesis:

end;

hence Sum x = 0 by A35, A36, CARD_1:27; :: thesis:

end;

assume dom (f | ((dom f) /\ (eq_dom (f,0)))) <> {} ; :: thesis:
hence integral' (M,(f | (eq_dom (f,0)))) = 0 by A3, A26, A34, Def14; :: thesis:

end;

hence integral' (M,(f | (eq_dom (f,0)))) = 0 by A3, Def14; :: thesis: