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,REAL st f in L1_Functions M & Integral M,(abs f) = 0 holds
f a.e.= X --> 0 ,M
let S be SigmaField of X; :: thesis: for M being sigma_Measure of S
for f being PartFunc of X,REAL st f in L1_Functions M & Integral M,(abs f) = 0 holds
f a.e.= X --> 0 ,M
let M be sigma_Measure of S; :: thesis: for f being PartFunc of X,REAL st f in L1_Functions M & Integral M,(abs f) = 0 holds
f a.e.= X --> 0 ,M
let f be PartFunc of X,REAL ; :: thesis: ( f in L1_Functions M & Integral M,(abs f) = 0 implies f a.e.= X --> 0 ,M )
assume A1:
( f in L1_Functions M & Integral M,(abs f) = 0 )
; :: thesis: f a.e.= X --> 0 ,M
then consider f1 being PartFunc of X,REAL such that
A2:
( f = f1 & ex ND being Element of S st
( M . ND = 0 & dom f1 = ND ` & f1 is_integrable_on M ) )
;
consider ND being Element of S such that
A3:
( M . ND = 0 & dom f1 = ND ` & f1 is_integrable_on M )
by A2;
A4:
abs f is_integrable_on M
by Lm15, A2;
then
R_EAL (abs f) is_integrable_on M
by MESFUNC6:def 9;
then consider E being Element of S such that
A5:
( E = dom (R_EAL (abs f)) & R_EAL (abs f) is_measurable_on E )
by MESFUNC5:def 17;
A6:
( E = dom (abs f) & abs f is_measurable_on E )
by A5, MESFUNC6:def 6;
then A7:
abs f is nonnegative
by MESFUNC6:52;
defpred S1[ Element of NAT , set ] means ( $2 = great_eq_dom (abs f),(1 / ($1 + 1)) & M . $2 = 0 );
A8:
now let n be
Element of
NAT ;
:: thesis: ex B being Element of S st S1[n,B]set r = 1
/ (n + 1);
reconsider Br =
E /\ (great_eq_dom (abs f),(1 / (n + 1))) as
Element of
S by A6, MESFUNC6:13;
A9:
Integral M,
((abs f) | Br) <= Integral M,
((abs f) | E)
by A6, A7, MESFUNC6:87, XBOOLE_1:17;
set g =
Br --> (1 / (n + 1));
A10:
dom (Br --> (1 / (n + 1))) = Br
by FUNCT_2:def 1;
A11:
for
x being
set st
x in dom (Br --> (1 / (n + 1))) holds
(Br --> (1 / (n + 1))) . x = 1
/ (n + 1)
by FUNCOP_1:13;
reconsider g =
Br --> (1 / (n + 1)) as
PartFunc of
X,
REAL by RELSET_1:17;
A12:
g is_measurable_on Br
by A10, A11, Lm261a;
for
x being
set st
x in dom g holds
0 <= g . x
by A11;
then A13:
g is
nonnegative
by MESFUNC6:52;
dom ((abs f) | Br) =
(dom (abs f)) /\ Br
by RELAT_1:90
.=
Br
by A5, XBOOLE_1:17, XBOOLE_1:28
;
then A14:
(
dom g = dom ((abs f) | Br) &
(abs f) | Br is_integrable_on M )
by A4, FUNCT_2:def 1, MESFUNC6:91;
for
x being
set st
x in great_eq_dom (abs f),
(1 / (n + 1)) holds
x in dom (abs f)
by MESFUNC6:6;
then A15:
great_eq_dom (abs f),
(1 / (n + 1)) c= E
by A5, TARSKI:def 3;
then A19:
Integral M,
(abs g) <= Integral M,
((abs f) | Br)
by A12, A14, A10, MESFUNC6:96;
A20:
dom (abs g) = dom g
by VALUED_1:def 11;
then A23:
Integral M,
g <= Integral M,
((abs f) | Br)
by A19, A20, PARTFUN1:34;
0 <= Integral M,
g
by A10, A11, Lm261a, A13, MESFUNC6:84;
then A25:
Integral M,
g = 0
by A9, A5, A1, A23, RELAT_1:97;
(
0 <= 1
/ (n + 1) &
dom g in S & 1
/ (n + 1) is
Real & ( for
x being
set st
x in dom g holds
g . x = 1
/ (n + 1) ) )
by A10, FUNCOP_1:13;
then
(R_EAL (1 / (n + 1))) * (M . Br) = 0
by A10, A25, MESFUNC6:97;
then
M . Br = 0
;
hence
ex
B being
Element of
S st
S1[
n,
B]
by A15, XBOOLE_1:28;
:: thesis: verum end;
consider F being Function of NAT ,S such that
A29:
for n being Element of NAT holds S1[n,F . n]
from FUNCT_2:sch 3(A8);
then A31:
{ x where x is Element of X : ( x in dom f & f . x <> 0 ) } c= union (rng F)
by TARSKI:def 3;
then
union (rng F) c= { x where x is Element of X : ( x in dom f & f . x <> 0 ) }
by TARSKI:def 3;
then A30:
{ x where x is Element of X : ( x in dom f & f . x <> 0 ) } = union (rng F)
by A31, XBOOLE_0:def 10;
then A48:
union (rng F) is measure_zero of M
by MEASURE2:16;
reconsider EQ = (union (rng F)) \/ ND as Element of S ;
ND is measure_zero of M
by A3, MEASURE1:def 13;
then
EQ is measure_zero of M
by A48, MEASURE1:70;
then A49:
M . EQ = 0
by MEASURE1:def 13;
set g = X --> 0 ;
A51:
EQ ` = (ND ` ) /\ ((union (rng F)) ` )
by XBOOLE_1:53;
then A51a:
EQ ` c= dom f
by A2, A3, XBOOLE_1:17;
A52:
dom (f | (EQ ` )) = EQ `
by A2, A3, A51, RELAT_1:91, XBOOLE_1:17;
dom (X --> 0 ) = X
by FUNCOP_1:19;
then A53:
dom ((X --> 0 ) | (EQ ` )) = EQ `
by RELAT_1:91;
then
f | (EQ ` ) = (X --> 0 ) | (EQ ` )
by A51a, A53, FUNCT_1:9, RELAT_1:91;
hence
f a.e.= X --> 0 ,M
by Def2, A49; :: thesis: verum