let X be non empty set ; :: thesis: for S being SigmaField of X
for M being sigma_Measure of S
for f being PartFunc of X,ExtREAL st f is_simple_func_in S holds
( max+ f is_simple_func_in S & max- f is_simple_func_in S )
let S be SigmaField of X; :: thesis: for M being sigma_Measure of S
for f being PartFunc of X,ExtREAL st f is_simple_func_in S holds
( max+ f is_simple_func_in S & max- f is_simple_func_in S )
let M be sigma_Measure of S; :: thesis: for f being PartFunc of X,ExtREAL st f is_simple_func_in S holds
( max+ f is_simple_func_in S & max- f is_simple_func_in S )
let f be PartFunc of X,ExtREAL; :: thesis: ( f is_simple_func_in S implies ( max+ f is_simple_func_in S & max- f is_simple_func_in S ) )
assume A1: f is_simple_func_in S ; :: thesis: ( max+ f is_simple_func_in S & max- f is_simple_func_in S )
then A2: f is real-valued by MESFUNC2:def 4;
consider F being Finite_Sep_Sequence of S such that
A3: dom f = union (rng F) and
A4: for n being Nat
for x, y being Element of X st n in dom F & x in F . n & y in F . n holds
f . x = f . y by A1, MESFUNC2:def 4;
A5: dom (max+ f) = union (rng F) by A3, MESFUNC2:def 2;
for n being Nat
for x, y being Element of X st n in dom F & x in F . n & y in F . n holds
(max+ f) . x = (max+ f) . y

proof

let n be Nat; :: thesis: for x, y being Element of X st n in dom F & x in F . n & y in F . n holds
(max+ f) . x = (max+ f) . y
let x, y be Element of X; :: thesis: ( n in dom F & x in F . n & y in F . n implies (max+ f) . x = (max+ f) . y )
assume that
A6: n in dom F and
A7: x in F . n and
A8: y in F . n ; :: thesis: (max+ f) . x = (max+ f) . y
F . n in rng F by A6, FUNCT_1:3;
then ( x in dom (max+ f) & y in dom (max+ f) ) by A5, A7, A8, TARSKI:def 4;
then ( (max+ f) . x = max ((f . x),0) & (max+ f) . y = max ((f . y),0) ) by MESFUNC2:def 2;
hence (max+ f) . x = (max+ f) . y by A4, A6, A7, A8; :: thesis: verum

end;

hence max+ f is_simple_func_in S by A2, A5, Th2, MESFUNC2:def 4; :: thesis: max- f is_simple_func_in S
A9: dom (max- f) = union (rng F) by A3, MESFUNC2:def 3;
for n being Nat
for x, y being Element of X st n in dom F & x in F . n & y in F . n holds
(max- f) . x = (max- f) . y

proof

let n be Nat; :: thesis: for x, y being Element of X st n in dom F & x in F . n & y in F . n holds
(max- f) . x = (max- f) . y
let x, y be Element of X; :: thesis: ( n in dom F & x in F . n & y in F . n implies (max- f) . x = (max- f) . y )
assume that
A10: n in dom F and
A11: x in F . n and
A12: y in F . n ; :: thesis: (max- f) . x = (max- f) . y
F . n in rng F by A10, FUNCT_1:3;
then ( x in dom (max- f) & y in dom (max- f) ) by A9, A11, A12, TARSKI:def 4;
then ( (max- f) . x = max ((- (f . x)),0) & (max- f) . y = max ((- (f . y)),0) ) by MESFUNC2:def 3;
hence (max- f) . x = (max- f) . y by A4, A10, A11, A12; :: thesis: verum

end;

hence max- f is_simple_func_in S by A2, A9, Th2, MESFUNC2:def 4; :: thesis: verum