reserve X for set;

theorem Th19:
  for S being SigmaField of X, M being sigma_Measure of S, B1,B2
being set st B1 in S & B2 in S holds for C1,C2 being thin of M holds B1 \/ C1 =
  B2 \/ C2 implies M.B1 = M.B2
proof
  let S be SigmaField of X, M be sigma_Measure of S, B1,B2 be set;
  assume
A1: B1 in S & B2 in S;
  let C1,C2 be thin of M;
  assume
A2: B1 \/ C1 = B2 \/ C2;
  then
A3: B1 c= B2 \/ C2 by XBOOLE_1:7;
A4: B2 c= B1 \/ C1 by A2,XBOOLE_1:7;
  consider D1 being set such that
A5: D1 in S and
A6: C1 c= D1 and
A7: M.D1 = 0. by Def2;
A8: B1 \/ C1 c= B1 \/ D1 by A6,XBOOLE_1:9;
  consider D2 being set such that
A9: D2 in S and
A10: C2 c= D2 and
A11: M.D2 = 0. by Def2;
A12: B2 \/ C2 c= B2 \/ D2 by A10,XBOOLE_1:9;
  reconsider B1,B2,D1,D2 as Element of S by A1,A5,A9;
A13: M.(B1 \/ D1) <= M.B1 + M.D1 & M.B1 + M.D1 = M.B1 by A7,MEASURE1:33
,XXREAL_3:4;
  M.B2 <= M.(B1 \/ D1) by A4,A8,MEASURE1:31,XBOOLE_1:1;
  then
A14: M.B2 <= M.B1 by A13,XXREAL_0:2;
A15: M.(B2 \/ D2) <= M.B2 + M.D2 & M.B2 + M.D2 = M.B2 by A11,MEASURE1:33
,XXREAL_3:4;
  M.B1 <= M.(B2 \/ D2) by A3,A12,MEASURE1:31,XBOOLE_1:1;
  then M.B1 <= M.B2 by A15,XXREAL_0:2;
  hence thesis by A14,XXREAL_0:1;
end;
