# How to show a sequence is not uniformly integrable

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?

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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

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

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 $\mu\bigl((X\setminus B)\cap[n,n+1]\bigr)\ge1/2$.

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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