972201013
Question:
Can we release the conditios on the B.C.T. (thm 1.4) ?
Answer:
the answer is "no".
that is, the condictions of B.C.T. are sharp
(i) f_n's are not bounded
for positive integer n,
let f_n be defined on [0,1] by
f_n(x)={ n, if x is in [ 0,1/n ]
{
{ 0, if x is in ( 1/n,1 ]
then {f_n} is a sequence of Lebesgue integrable with unbounded
function on [0,1]
for 0<x≦1, we have f_n(x)→f(x)=0 as n→∞
where f(x)={ 0, if x is in (0,1]
{
{∞, is x=0
since ∫ f_n(x)dx = n * m([ 0,1/n ])+ 0 * m(( 1/n,1 ]) =1 for all n
[0,1]
therefore, ∫ f_n(x)dx 不趨近於 ∫ f(x)dx =0 as n→∞
[0,1] [0,1]
(ii)supp(f_n) contains in E , with m(E) = ∞
for positive integer n,
let f_n be defined on [0,∞) by
f_n(x)={ 1/n, if x is in [ 0,n ]
{
{ 0 , if x > n
for every x ≧0, we have f_n(x)→f(x)=0 as n→∞
since ∫ f_n(x)dx = 1 , for all n
[0,∞)
supp(f_n)=[0,n] →[0,∞) as n→∞
therefore , we have
∫ f_n(x)dx 不趨近於 ∫ f(x)dx = 0 as n→∞
[0,∞) [0,∞)
by the above two examples, we know that the conditions of
Bounded Convergence Theorem are sharp.

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