Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(X,\Omega,\mu)$ be a measure space. A sequence $f_n$ is said to be uniformly integrable if for every $\epsilon \gt 0$ there is a $\delta \gt 0$ such that for every measurable set $A$ with $\mu(A)\lt \delta$, $\int_A |f_n|~d\mu \lt \epsilon$, for every $n\in \mathbb{N}$.

A sequence is said to be tight if for every $\epsilon \gt 0$ there is a measurable set $B$ of finite measure such that $\int_{X\setminus B} |f_n|~d\mu \lt \epsilon $, for every $n\in \mathbb{N}$.

I claim the $f_n = n\cdot 1_{[0,1/n]}$ is not uniformly integrable and $g_n = 1_{[n,n+1]}$ is not tight.

Proof. Fix $\epsilon \gt 0$. Pick $n$ sufficiently large so that for every $\delta \gt 0$, $n\delta \gt 1/2.$ Then there is an $n$ such that $\int_A |f_n| \gt 1/2.$

Let $\mu(B)\lt \infty$. Suppose to the contrary that $g_n$ were tight. Then $\mu\left((X\setminus B)\cap [n,n+1]\right) \lt \epsilon$. If I can get that $\mu(B) = \infty$, then I would have a contradiction, but I don't see how.

Is what I have done above right?

share|improve this question
    
The subject is somewhat tricky, at least for me, so maybe this can help, it is a sort of an attempt to demonstrate what the uniform integrability is: websfog.blogspot.co.il/2012/09/probability-1.html –  user47756 Nov 1 '12 at 5:12
    
@robjohn I fail to see how your comment on the deleted answer concerns anybody else than the author of the answer. –  Did Nov 1 '12 at 8:42
add comment

1 Answer

up vote 1 down vote accepted

Your arguments are not correct. It seems you are mixing up or ignoring the quantifiers.


For your first claim:

You are trying to show that $\{f_n\}$ is not uniformly integrable. The definition of uniform integrablity states that something is true for all $\epsilon>0$. So if a sequence is not uniformly integrable, that something must be false for some $\epsilon>0$. Here, as it turns out, you can take $\epsilon=1/2$; this will be the $\epsilon$ that "doesn't work" in the definition.

So, set $\epsilon=1/2$. We then have to show that the following is not true:

$\ \ \ $There is a $\delta>0$ such that for every set $B$ with $\mu(B)<\delta$, we have $\int_B|f_n|<1/2$ for all $n$.

So, we have to show that, given $\delta>0$, there is a set with measure less that $\delta$ but with the integral of some $f_n$ over that set greater than or equal to $1/2$. Towards this end, you can do the following:

Let $\delta>0$. Choose $1/N<\delta$. Then $\mu([0,1/N])<\delta$, but $\int_{[0,1/N]} |f_N|=N\cdot{1\over N}=1$.

Since $1>1/2$, we are done.


For the second claim:

I would argue directly: "We will show $\{g_n\}$ is not tight".

The definition of tightness says that something is true for all $\epsilon>0$. So you have to show that that something is not true for some $\epsilon>0$. As it turns out you can take $\epsilon=1/2$; this will be the $\epsilon$ that "doesn't work" in the definition.

What "doesn't work" here is the following:

$\ \ \ \ $There is a set $B$ of finite measure with $\int_{X\setminus B} |g_n|<1/2$ for all $n$.

So we have to show the above statement is not true. So, we have to show that for any $B$ with finite measure, the following does not hold: $\int_{X\setminus B} |g_n|<1/2$ for all $n$.

So, let $\mu(B)$ be finite. $B$ is fixed now, and your task is to show that for some $n$, $\int_{X\setminus B} |g_n|\ge1/2$.

Once you've done this, you'll have your result.

A hint for this: since the measure of $B$ is finite there is an $n$ with $\mu\bigl((X\setminus B)\cap[n,n+1]\bigr)\ge1/2$.

share|improve this answer
    
I meant to say fix $\epsilon = 1/2.$ why do you have $\int_{[1,1/n]} |f_n|$? –  Jonjo Apr 4 '12 at 17:09
    
@Jonjo Sorry, that was a typo. I rewrote the answer; I hope it helps... –  David Mitra Apr 4 '12 at 17:35
    
Thanks. I think I got it now, but why $\mu(B\cap[n,n+1])\ge1/2$ and not $\mu((X\setminus B)\cap[n,n+1])\ge1/2$? –  Jonjo Apr 4 '12 at 18:18
    
@Jonjo Bleh, another "typo"... Thanks. –  David Mitra Apr 4 '12 at 18:24
    
So would any $n$ such the $\mu([n,n+1]) \gt 1/2$ suffice? –  Jonjo Apr 4 '12 at 18:39
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.