Corollary 5: Let $\{f_n\}$ be a sequence of nonnegative integrable functions on $E$. Then

$$\lim_{n \to \infty} \int_E f_n = 0 ~~~~~~(5)$$

if and only if

$$f_n \to 0 \mbox{ in measure on } E \mbox{ and } \{f_n\} \mbox{ is uniformly integrable and tight over } E ~~~~~(6)$$

Here's the part of the proof giving me trouble:

First assume (5). Corollary 2 tells us that $\{f_n\}$ is uniformly integrable and tight over $E$...

Here is corollary 2:

Let $\{h_n\}$ be a sequence of nonnegative integrable functions on $E$. Suppose that $h_n(x) \to 0$ for all most all $x \in E$. Then $$\int_E h_n \to 0 \mbox{ iff } \{h_n\} \mbox{ uniformly integrable and tight over } E$$

How can we invoke corollary 2 if $f_n$ isn't assumed to converge pointwise a.e. to $0$ on $E$?


Okay. I encountered some more difficulties in the second half of the proof:

To prove the converse, we argue by contradiction. Assume that (6) holds but (5) fails to hold. Then there is some $\epsilon_0 > 0$ and a subsequence $\{f_{n_k}\}$ for which $$\int_E f_{n_k} \ge \epsilon_0 \mbox{ for all } k$$ However, by theorem 4, a subsequence $\{f_{n_k}\}$ converges to $f \equiv 0$ pointwise almost everywhere on $E$ and this subsequence is uniformly integrable and tight so that, by the Vitali Convergence Theorem, we arrive at a contradiction to the existence of the above $\epsilon_0$. This completes the proof.

First, why does $\int_E f_{n_k} \ge \epsilon_0$ hold for all $k$? If I've properly negated the definition of the convergence of a sequence, we should actually have that there exists an $\epsilon > 0$ such that for every $N \in \Bbb{N}$ there exists an $k \ge N$ such that $\int_E f_{n_k} \ge \epsilon_0$.

Second, assuming that (5) fails certainly does give us a subsequence such that blah blah holds; and theorem 4 gives us a subsequence such that blah blah holds. But what reason is there for thinking that these subsequences are the same?

  • $\begingroup$ There is an erratum to this Corollary, though I don't immediately see that it resolves your questions. $\endgroup$
    – aduh
    Apr 2, 2018 at 14:38
  • 1
    $\begingroup$ As to questions in Edit, I don't see what the problem is with $\int_E f_{n_k} \geq \epsilon_0$. For each $n$, just define $n_k$ to be any $k \geq n$ such that the desired inequality holds. For the second part, the authors don't assume the subsequences are the same. They write, "a subsequence of $\{f_{n_k}\}$ converges...," that subsequence being a subsubsequence of $\{f_n\}$. $\endgroup$
    – aduh
    Apr 2, 2018 at 14:46
  • $\begingroup$ @aduh I'm afraid I don't understanding your reasoning. We are given the sequence in virtue of assuming (5) false. So we don't have any control over its form/nature, so how can modify it so that the inequality holds for every $k$? $\endgroup$
    – user193319
    Apr 13, 2018 at 14:41
  • $\begingroup$ The negation of (5) is: $\exists \epsilon_0>0, \forall n, \exists n_k \geq n$ such that $\int_E f_{n_k} \geq \epsilon_0$. So $\{f_{n_k}\}$ is the desired subsequence. I still don't see what the issue is here. $\endgroup$
    – aduh
    Apr 13, 2018 at 14:55
  • 1
    $\begingroup$ There’s nothing wrong with my negation. It’s a mere notational variation of what you wrote that makes the relevant relation to subsequences salient. You seem to be confused about some basic definitions. I’m not sure what else I can say to help you with this. You should just review and think carefully about the definitions of subsequences, limits, etc. $\endgroup$
    – aduh
    Apr 13, 2018 at 18:39

1 Answer 1


I responded to your EDIT in the comments so will confine this answer to the main question. However, if you have further questions about the EDIT, I'd be glad to try to answer them. (Sorry if my last comment seems rude. It was written in a hurry.)

Now, as to your main question, I think Royden's exposition here isn't great, but the result still holds.

Here, I think, is the most charitable way to read the text. Section 5.1 extends the results in the preceding section, 4.6, by proving a Vitali Convergence Theorem for sets with infinite measure. Section 5.2 introduces the concept of convergence in measure and extends some of the results from 4.6 and 5.1. To understand the result you ask about, it's helpful, then, to view it as an analogue of the results that precede it in sections 4.6 and 5.1. In particular, if you inspect the proof of Theorem 26 in Section 4.6, you will see that $\lim_{n \to \infty}\int_E h_n = 0$ implies $\{h_n\}$ is uniformly integrable on $E$; the assumption that $h_n \to 0$ a.e. is not used in this direction of the proof. Corollary 2 in Section 5.1, the proof of which is left as an exercise, is a straightforward extension of Theorem 26.

The part of Corollary 2 that is relevant to your question is: $\lim_{n \to \infty}\int_E h_n = 0$ implies $\{ h_n\}$ is uniformly integrable and tight. To get uniform integrability, the proof of Theorem 26 applies. To get tightness, the same idea works. To wit, with $\epsilon > 0$ given, $\lim_{n \to \infty}\int_E h_n = 0$ implies that $\int_{E_0} h_n < \epsilon$ for all $n >N$ and all $E_0 \subset E$ because the $\{h_n\}$ are non-negative. If $n \leq N$, then the integrability of $h_1,...,h_N$ implies that there is a set $A$ of finite measure such that $$\int_{E \setminus A} h_j < \epsilon, \ \ 1 \leq j \leq N.$$ This is Proposition 1, Section 5.1, extended to a finite collection of integrable functions. Because $E \setminus A \subset E$, it follows that $\{h_n\}$ is tight.

To sum up, the assumption that $f_n \to 0$ a.e. is not needed in the direction of the proof that you ask about, but I do think that the authors could have made this clearer somehow.

  • $\begingroup$ Ah! Dang it! I worked on corollary 2, but I suppose enough time elapsed between proving corollary 2 and proving corollary 5 that I forgot that one direction of corollary 2 doesn't use pointwise convergence. By the way, I don't you think you were being rude, so no worries. Thanks for the help! $\endgroup$
    – user193319
    Apr 26, 2018 at 16:26
  • $\begingroup$ By the way, if you're interested in receiving another 50 points, I bounty on this question $\endgroup$
    – user193319
    Apr 26, 2018 at 16:29

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