reserve X for set,
  F for Field_Subset of X,
  M for Measure of F,
  A,B for Subset of X,
  Sets for SetSequence of X,
  seq,seq1,seq2 for ExtREAL_sequence,
  n,k for Nat;
reserve FSets for Set_Sequence of F,
  CA for Covering of A,F;
reserve Cvr for Covering of Sets,F;
reserve C for C_Measure of X;

theorem
  for X being non empty set, F being Field_Subset of X, m being Measure
  of F st (ex M be sigma_Measure of sigma F st M is_extension_of m) holds m is
  completely-additive
proof
  let X be non empty set, F be Field_Subset of X, m be Measure of F;
  given M be sigma_Measure of sigma F such that
A1: M is_extension_of m;
A2: F c= sigma F by PROB_1:def 9;
  for Aseq being Sep_Sequence of F st union rng Aseq in F holds SUM(m*Aseq
  ) = m.(union rng Aseq)
  proof
    let Aseq be Sep_Sequence of F;
    reconsider Bseq = Aseq as sequence of sigma F by A2,FUNCT_2:7;
    reconsider Bseq as Sep_Sequence of sigma F;
A3: now
      let n be object;
      assume n in NAT;
      then reconsider n9 = n as Element of NAT;
      (M*Bseq).n = M.(Bseq.n9) by FUNCT_2:15;
      then (M*Bseq).n = m.(Aseq.n9) by A1;
      hence (M*Bseq).n = (m*Aseq).n by FUNCT_2:15;
    end;
    assume union rng Aseq in F;
    then m.(union rng Aseq) = M.(union rng Aseq) by A1;
    then m.(union rng Aseq) = SUM(M*Bseq) by MEASURE1:def 6;
    hence thesis by A3,FUNCT_2:12;
  end;
  hence m is completely-additive;
end;
