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

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.

## 2 Answers

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

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.

• oh wow I see what you are saying, thank you – Jordan Reed Nov 16 '19 at 20:41