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I know I keep asking similar sorts of questions, but I want to understand exactly what is going on here. So, I will list a full definition and then the problem. (Basically, I want to prove something using Vitali's theorem)

A sequence of functions $\{f_n\} \in L_1(\mu)$ is called uniformly integrable if for every $\epsilon >0$ there is a $\delta > 0$ such that for all $E$ with $\mu(E) < \delta$, then $|\int_E{f_nd\mu}| < \epsilon$ for all $n$.

The problem is as follows : Suppose that $(X,M,\mu)$ is a measure space, with $\mu(X) < \infty$ and $f_n \in L_1(\mu)$ for all n. Also for some $p>1$ we have that $||f_n||_p < K$, where $K$ is a positive real number. Prove that $\{f_n\}$ is uniformly integrable.

So the problem is asking for asking to show a $L_p$ bounded sequence is uniformly integrable. Now, this only assumes the existence of SOME $p >1$, so I can't imagine that it's that important. Also, I thought it was strange why we need to assume that $f_n \in L_1$. Don't we get this by the boundedness condition and also the fact that the measure space is finite?

In any case, I'm lost. Any ideas?

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  • $\begingroup$ "Also, I thought it was strange why we need to assume that $f_n\in L_1$. Don't we get this by the boundedness condition and also the fact that the measure space is finite?" The condition $f_n\in L_1$ is redundant, but there is no harm in mentioning it and since uniform integrability is only defined for functions in $L_1$ it seems prudent. $\endgroup$
    – Alex Becker
    Aug 20, 2012 at 5:11
  • $\begingroup$ Thank you Alex, I have corrected the definition including the $\delta$ part. $\endgroup$
    – James
    Aug 20, 2012 at 5:25
  • $\begingroup$ @AlexBecker and James: The condition that $f_n\in L^1$ is NOT redundant and in fact this condition should be replaced by the stronger condition that $(f_n)$ is bounded in $L^1$: assume that $\mu$ is discrete and that every atom has mass $\geqslant\alpha$, then the other condition says nothing when $\delta\lt\alpha$, hence every family of random variables would be uniformly integrable without this additional condition. Of course, when $\mu$ has no atom, all this is implied by the $\varepsilon-\delta$ condition. $\endgroup$
    – Did
    Aug 20, 2012 at 8:15
  • $\begingroup$ Does it hold for $p=\infty$ too? $\endgroup$
    – C. Bishop
    Apr 3, 2021 at 9:33

1 Answer 1

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Suppose $\{f_n\}$ is not uniformly integrable, so we have some $\epsilon>0$ such that there exists a sequence of sets $\{E_k\}$ with $\mu(E_k)\to 0$ and for all $k$, there is some $n$ such that $\left|\int_{E_k} f_nd\mu\right|\geq\epsilon$. Then $$\epsilon\leq\int_{E_k}|f_n|d\mu\leq\left(\int_{E_k}|f_n|^{p}d\mu\right)^{1/p}\left(\int_{E_k}1^{q}d\mu\right)^{1/q}=\mu(E_k)^{1/q}\left(\int_{E_k}|f_n|^{p}d\mu\right)^{1/p}\leq K\mu(E_k)^{1/q}$$ thanks to Holder's inequality where $1/p+1/q=1$. Thus $K\geq 1/\mu(E_k)^{1/q}\to \infty$, a contradiction.

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  • $\begingroup$ Dang it! I knew I was on the right track... all I had to do was stare at it a little longer. Oh well... good answer and thanks! $\endgroup$
    – James
    Aug 20, 2012 at 5:38
  • $\begingroup$ @James No problem. BTW, I recommend you register here at math.SE so that you can use your reputation to gain privileges and keep better track of your questions/answers. $\endgroup$
    – Alex Becker
    Aug 20, 2012 at 5:41
  • $\begingroup$ By the way, it's much more straightforward to prove the sequence is uniformly integrable directly: By Holder, $\int_E|f_n|\,d\mu \le C\mu(E)^{1/q} < \epsilon$ whenever $\mu(E) < (\epsilon/C)^q$. $\endgroup$
    – Alex Ortiz
    Dec 3, 2017 at 20:55
  • $\begingroup$ Does it hold for $p=\infty$ too? $\endgroup$
    – C. Bishop
    Apr 3, 2021 at 9:33

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