
theorem Th35:
for X be set, S be SigmaField of X, M be sigma_Measure of S, E be Element of S
 holds M.E = (COM M).E
proof
    let X be set, S be SigmaField of X, M be sigma_Measure of S,
    E be Element of S;
    reconsider E0 = {} as thin of M by Th34;
    (COM M).(E \/ E0) = M.E by MEASURE3:def 5;
    hence thesis;
end;
