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.
 A: As David Mitra showed, the result does not hold with this definition of uniform integrability. The main reason is that it is not adapted to infinite measure spaces: if $f$ is a bounded function, then $\{f\}$ is a uniformly integrable (UI) family but $f$ may be not integrable it self. 
We rather use the definition 

The family $\{f_i,i\in I\}$ is uniformly integrable if for each $\varepsilon\gt 0$, there is an integrable function $g$ such that 
  $$\sup_{i\in I}\int_X(|f_i|-g)^+\mathrm d\mu\lt\varepsilon.$$

With this definition, we have the result 

Let $(X,\mathcal A,\mu)$ be a measure space, $\{f_n,n\geqslant 1\}$ a UI sequence and $f$ such that $f_n\to f$ in measure. Then $f_n\to f\in\mathbb L^1$. 

Proof: Considering $B_{n,k}:=\{|f_n|\geqslant k^{-1}\}$, that we rearrange into an increasing sequence $(A_k)_{k\geqslant 1}$, we have by an approximation argument that 
$$\lim_{k\to \infty}\sup_n\int_{X\setminus A_k}|f_n|\mathrm d\mu=0.$$
Fix $(f_{n'})$ a subsequence of $(f_n)$. Using a diagonal argument, we can extract a further subsequence $(f_{n''})$ which converges almost everywhere to $f$. We thus have by Fatou's lemma: $\int_{X\setminus A}|f|\mathrm d\mu=0$ where $A:=\bigcup_kA_k$. We then conclude by dominated convergence.
A: 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 $A$.
