Author |
Topic: lebesgue Integral (Read 553 times) |
|
trusure
Newbie
Gender: 
Posts: 24
|
 |
lebesgue Integral
« on: Oct 9th, 2009, 8:58pm » |
 Quote  Modify
|
I need to prove that the function
F(x)=integral_infinity to x ( f(u) du )
is continiuous if f is lebesgue integrable.
I need to prove it using the Monotone convergence theorem, of course I can prove it by other ways,,, but I need to construct a sequence of nonnegative non-decreasing integrable function {h_n} on R (real numbers), then apply the result which is
limit (integral h_n) = integral (limit h_n)
??
|
|
 IP Logged |
|
|
|
Obob
Senior Riddler
Gender: 
Posts: 489
|
|
Re: lebesgue Integral
« Reply #1 on: Oct 9th, 2009, 10:26pm » |
Quote Modify
|
Let x_n be an increasing monotone sequence converging to x. Let h_n(y) = f(y) if y<=x_n, and h_n(y) = 0 otherwise. Apply MCT to this.
|
|
IP Logged |
|
|
|
trusure
Newbie
Gender:
Posts: 24
|
|
Re: lebesgue Integral
« Reply #2 on: Oct 10th, 2009, 7:33am » |
Quote Modify
|
you are right, but I think that this definition for continuity should apply for any sequence x_n converges to x in general.
I did it for an increasing sequence x_n , but is that enough ??!!
|
|
IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: lebesgue Integral
« Reply #3 on: Oct 10th, 2009, 7:18pm » |
Quote Modify
|
You also need to check decreasing sequences; checking both increasing and decreasing sequences is enough to check continuity (a fact which you should be able to prove easily).
If your functions are allowed to have negative values, then the dominated convergence theorem gives a more straightforward argument than the MCT.
I think a better arugment is the following, though: if f is in L1, then for every epsilon there exists a delta such that if S is a set of measure at most delta, then the integral of |f| over S is at most epsilon. This fact immediately implies the result you want.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print