Let $S$ be a set of measurable functions $E \to [- \infty, \infty]$. We say that $S$ is uniformly integrable iff for every $\epsilon>0$ there exists a $\delta >0$ such that, for any $f \in S$ and any $A \subseteq E$ with $m(A)< \delta$, we have $\int_A|f|< \epsilon.$ A sequence $\{f_n\}$ of measurable functions $E \to [- \infty, \infty]$ is said to be uniformly integrable iff the set $\{f_1,f_2,...\}$ is.
(I know that uniform integrability has a special meaning in probability theory, but I don't care about probability right now! I'm doing real analysis. By the way, we are working in the real numbrs, with the Lebsgue measure, with the Lebesgue integral.)
Let $\{g_n\}$ and $\{h_n\}$ be sequences of uniformly integrable functions $E \to [-\infty, \infty]$. I know that $\{ \alpha g_n+ \beta h_n\}$ must be uniformly integrable for any choice of $\alpha, \beta \in \mathbb{R}$. (At least if the sum is actually defined and we aren't adding $\infty+- \infty$ or anything like that.)
My question is, what about $\{g_nh_n\}$? I am pretty sure that this doesn't have to be uniformly integrable in general. For instance, I see we can take $E=(0,1]$ and each $g_n(x)=h_n=1/\sqrt x$, but then $g_nh_n(x)=1/x$ is not integrable, hence the set $\{g_1h_1,g_2h_2,...\}=\{g_1h_1\}$ is not uniformly integrable.
So what extra assumptions would we need in order to guarantee that the sequence $\{g_nh_n\}$ is uniformly integrable?
I was thinking that it might be enough to assume $\int_E |g_nh_n|< \infty$ for each $n$. But when I try to work out a proof I get stuck because it may not be true that $\int_A|g_nh_n| \leq (\int_A|g_n|)(\int_A |h_n|)$. (See my previous question Looking for an inequality relating $\int_Efg$ to the integrals $\int_Ef$ and $\int_Eg$). Is there a quick counterexample to show that $\int_E |g_nh_n|< \infty$ might not be strong enough to guarantee $g_nh_n$?