Recent content by onthetopo

I'm sorry this is not pertaining to my question at all. I am not asking a computer programming question. Dynamic programming is a field in mathematics.

Hi, I am proficient in standard dynamic programming techniques. In the standard textbook reference, the state variable and the control variable are separate entities. However, I have seen examples in economics, in which a single variable, let's say consumption, is both a state variable and a...

if x is in (0,1/n), then f_n(x)=n; and if x is in (-1/n,0), then f_n(x)=-n., and if x is in neither, f_n(x)=0 thus, |f_n(x)|<|n| for all x on (-1,1) , but|n| is unbounded as n goes to infinity we cannot find a g(x) in L1 such that |fn|<|g| for all n and all x in (-1,1).

why do we need gn->g pointwise |gn|<|fn|+|hn|<|f|+|h|+2epsilon thus |gn| converges to |f|+|h|

ah! so it's dominated by |n|, which is not in L1, so we can;t use DCT at all

Wonderful, How did you prove this? if too long, what reference book did you find this theorem?

To use DCT, we need to show there exists a g such that g->gn but this is true, since |gn|<|fn|+|hn|<|f|+|h|+2epsilon thus |gn| converges to |f|+|h| Now, suppose the space is complete, then every absolutely convergent sequence is convergent, thus there exists a g such that gn->g is this right?

If x is in (0,1/n), then fn(x)<n*1/n-0=1 If x is in (-1/n,0), then fn(x)>0-n*1/n=-1 If x is in neither , then fn(x)=0 Thus |fn(x)| is dominated by |g| where g(x)=1, I really can't see what's wrong?

The question should include fn, hn, f,h integreable then max(|fn|,|hn|) is also integreable, then |gn|<x(|fn|,|hn|), thus gn is in L1, But still it doesn't say "integral gn" converges to "integral g" for some g

The question is as is, there is no mention of sigma finiteness. My attempt at the solution could be completely wrong.

Can someone please tell me if what i did is right?

E is a measurable set, let x be the characteristic function, then |f|xE is in L1 |g|xEc is in L1 and |f|xE+|g|xEc=|min(f,g)| is in L1 thus min (f,g) is in L1 QED please tell me this is right, exam is next week, thanks

Sorry , you are right. Ideally, if fn, gn, and hn are all nonnegative, we can use comparison theorem to show that hn is in L1. But how about the case that fn, gn, hn are simply any measurable functions?

Can we say that \int fn(x)=\int n\chi_{(0,n^{-1})}(x) +\int -n\chi_{(-n^{-1},0)}(x) -1<x<1 (**) I think the above is the crucial step. What exactly do we need for the above to hold? I guess as long as each term on the right is integreable? Assume (**) holds, then let u be the lebesgue...

It would be helpful if there is a sandwich theorem for lebesgue integreal. Does it exist? Ie. if fn<=hn<=gn on measure space (X,S,u) and fn and hn are integreable and measurable , then g must also be in L1. A related claim is that if fn-> f and hn->h, does gn also necessarily converge to some g?