
theorem
for I,J,K be non empty closed_interval Subset of REAL,
  g be PartFunc of [:[:REAL,REAL:],REAL:],REAL st [:[:I,J:],K:] = dom g
holds
  Integral(L-Meas,Integral1(Prod_Measure(L-Meas,L-Meas),R_EAL g)|(REAL\K)) = 0
proof
    let I,J,K be non empty closed_interval Subset of REAL,
    g be PartFunc of [:[:REAL,REAL:],REAL:],REAL;
    assume
A1: [:[:I,J:],K:] = dom g;

    set Rf = R_EAL g;
    set F1 = Integral1(Prod_Measure(L-Meas,L-Meas),Rf);

A2: dom Rf = [:[:I,J:],K:] by A1,MESFUNC5:def 7;
A3: dom F1 = REAL by FUNCT_2:def 1;

A4: K is Element of L-Field by MEASUR10:5,MEASUR12:75;

    set NK = REAL \ K;
    REAL in L-Field by PROB_1:5; then
A5: NK is Element of L-Field by A4,PROB_1:6;

    reconsider RL1 = F1|NK  as PartFunc of REAL,ExtREAL;

    now let z be Element of REAL;
     assume
A6:  z in dom RL1; then
     z in REAL & not z in K by A3,XBOOLE_0:def 5; then
A7:  dom ProjPMap2(Rf,z) = {} by A2,MESFUN16:26;

A8: RL1.z = F1.z by A6,FUNCT_1:47;
     F1.z = Integral(Prod_Measure(L-Meas,L-Meas),ProjPMap2(Rf,z))
       by MESFUN12:def 7;
     hence RL1.z = 0 by A7,A8,MESFUN16:1;
    end;
    hence thesis by A3,A5,MESFUN12:57;
end;
