1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.