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?
Best Answer
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.