reserve i,j,k,n,n1,n2,m for Nat;
reserve a,r,x,y for Real;
reserve A for non empty closed_interval Subset of REAL;
reserve C for non empty set;
reserve X for set;

theorem
  for f being Function of A,REAL st f|A is bounded & f is integrable
  holds abs(f) is integrable & |.integral(f).|<=integral(abs(f))
proof
  let f be Function of A,REAL;
  assume that
A1: f|A is bounded and
A2: f is integrable;
A3: max-(f) is integrable by A1,A2,Th22;
A4: (max+f)|A is bounded_above by A1,Th14;
A5: (max-f)|A is bounded_above by A1,Th16;
A6: max+(f) is total by Th13;
A7: (max-f)|A is bounded_below by Th17;
A8: (max+f)|A is bounded_below by Th15;
A9: max-(f) is total by Th13;
A10: max+(f) is integrable by A1,A2,Th20;
  then max+(f) + max-(f) is integrable by A6,A9,A4,A8,A5,A7,A3,INTEGRA1:57;
  hence abs(f) is integrable by RFUNCT_3:34;
A11: integral(max+(f))-integral(max-(f)) = integral(max+(f)-max-(f)) by A6,A9
,A4,A8,A5,A7,A10,A3,INTEGRA2:33
    .= integral(f) by RFUNCT_3:34;
A12: integral(max+(f))+integral(max-(f)) = integral(max+(f)+max-(f)) by A6,A9
,A4,A8,A5,A7,A10,A3,INTEGRA1:57
    .= integral(abs(f)) by RFUNCT_3:34;
  then
A13: integral(abs(f))-integral(f) =2*integral(max-(f)) by A11;
  for x st x in A holds (max-(f)).x >= 0 by RFUNCT_3:40;
  then integral(max-(f))>=0 by A9,A5,A7,INTEGRA2:32;
  then
A14: integral(abs(f))>=integral(f) by A13,XREAL_1:49;
A15: integral(abs(f))+integral(f) =2*integral(max+(f)) by A12,A11;
  for x st x in A holds (max+(f)).x >= 0 by RFUNCT_3:37;
  then integral(max+(f))>=0 by A6,A4,A8,INTEGRA2:32;
  then -integral(abs(f))<=integral(f) by A15,XREAL_1:61;
  hence thesis by A14,ABSVALUE:5;
end;
