$\lim_{n \to \infty} \int_E h_n = 0 \iff \{h_n \} $ is uniformly intergrable over $E$.

Question

In Royden's Real Analysis textbook, Theorem 26 states: Suppose $\{h_n \} $ is a sequence of non-negative integrable functions that converge pointwise a.e. on $E$ to $h = 0$. Then $$ \lim_{n \to \infty} \int_E h_n = 0 \iff \{h_n \} \text{ is uniformly intergrable over } E$$.

Here $E$ is a set of finite measure.

Evidently the theorem is false without the assumption that the $h_n$'s are non-negative. My question is why is this the case. I have read the proof that Royden provides for this theorem and none of it seems to rely upon the fact that the $h_n$'s are non -negative.

Any help would be highly appreciated.

Answer 1

Hint: Consider $$h_n(x):= \begin{cases} n & \text{if $x \in \left[0,\frac{1}{n}\right)$,} \\ -n & \text{if $x \in \left[\frac{1}{n},\frac{2}{n}\right)$,} \\ 0 & \text{otherwise.} \end{cases}$$

Answer 2

Fix $\varepsilon > 0$. Let $A \subset E$. For big enough $n$ $$ \int_A h_n = \int_A |h_n| < \varepsilon $$

That's the place where non-negativeness was used. If $h_n$ weren't non-negative, we had only $$ \int_A h_n < \varepsilon, $$ not $$ \int_A |h_n| < \varepsilon, $$ which is used in definition of uniformly integrable family of functions.