Does uniform integrability plus convergence in measure imply convergence in $L^1$? 
Does uniform integrability plus convergence in measure imply convergence in $L^1$?

I know this holds on a probability space. Does it hold on a general measure space? I have tried googling. It returned very few results on UI on measure spaces, and none of them mentioned a result like the one in the title. This comes from a discussion about another question.
The proof i have seen for a probability space breaks for a general measure space.
By UI, i mean
$\sup_{f}\int_{|f|>h} |f|d\mu $ goes to $0$ as $h$ goes to infinity.

On $\Bbb R$ with Lebesgue measure, the sequence $(f_n)$ defined by $f_n={1\over n}\cdot \chi_{[n,2n]}$ for each $n$ would furnish a counterexample. Here, $\chi_A$ is the indicator function on the set $A$.
