This corresponds to Exercise $3.11$(b) in Folland's Real Analysis Book. I was hoping someone could have a look at my proof. Specifically, I wanted to make sure the final paragraph of my argument holds.

The definition of uniform integrability provided is as follows: $\{f_\alpha\}_{\alpha \in A}$ is uniformly integrable if $\forall \epsilon>0$, $\exists \delta > 0$ such that $|\int_E f_\alpha d\mu| < \epsilon$ for all $\alpha \in A$ whenever $\mu(E) < \delta$ for any measurable $E$.

The problem gives us a sequence $f_n \to f$ in $L^1$. We are to show the entire sequence is uniformly integrable. Here's my attempt:

Set any $n \in \mathbb{N}. |f_n| = |f_n - f_m + f_m - f + f| \leq |f_n -f_m| + |f_m - f| + |f|$.

$\implies \int_E |f_n| d\mu \leq \int_E |f_n -f_m| d\mu + \int_E |f_m - f| d\mu + \int_E |f| d\mu.$

$f \in {L}^1 \implies f$ is uniformly integrable. $(f_n)_{n \in \mathbb{N}}$ is a convergent sequence in $L^1$ suggests that $m \to \infty$, we have:

$\int_E |f_n -f_m| d\mu \to 0$ and $\int_E |f_m - f| d\mu \to 0$.

Given $\epsilon > 0$ chosen and large enough $m$, whenever $\mu(E) < \delta$, (the $\delta > 0$ based on uniform integrability of $f$) we have:

$$ \int_E |f_n| d\mu \leq \int_E |f_n -f_m| d\mu + \int_E |f_m - f| d\mu + \int_E |f| d\mu < 3 \frac{\epsilon}{3} = \epsilon $$

This holds for all $n \in \mathbb{N}$ ($\because n$ was arbitrary) $\implies$ bounding holds for the entire sequence $(f_n)_{n \in \mathbb{N}}$, in other words, implying that $(f_n)_{n \in \mathbb{N}}$ is uniformly integrable. For greater detail, we could alternatively say that using the fact that is a Cauchy sequence, this holds for all $l \geq n$, which would leave us with the finite collection $(f_i)_{1 \leq i \leq n-1}$ to consider. To this, we could apply the procedure used to establish that a finite collection of $L^1$ functions is uniformly integrable and get the result.

UPDATE As very kindly pointed out by Friedrich Philipp, the above proof was founded on an incorrect premise (detailed in comments). I am leaving the proof I settled on eventually here for reference.

Fix $\epsilon > 0$. There exists $N_{\epsilon} \in \mathbb{N}$ such that for all $k > N_{\epsilon}$, by the fact of $L^1$ convergence, we have: $|\int_E f_k - f d\mu| \leq \int_E |f_k - f| d\mu < \frac{\epsilon}{2}$. considering $f \in L^1$ as a single element collection, by part (a), we can say that $f$ is uniformly integrable. That is, $\exists \delta_1 > 0$ such that $|\int_E f d\mu| < \frac{\epsilon}{2}$ whenever $\mu(E) < \delta_1$

Thus $\forall k > N_{\epsilon}$, we have:

$$|\int_E f_k d\mu| \leq \int_E |f_k - f| + |\int_E f d\mu| < \epsilon$$ whenever $\mu(E) < \delta_1$.

By part (a), we have that $(f_i)_{1 \leq i \leq N_{\epsilon}}$ is a finite collection in $L^1$ which is uniformly integrable, that is, $\exists \delta_0 > 0$ such that $|\int_E f_i d\mu | < \epsilon$ for all $i \in \{ 1, \dots, N_{\epsilon} \}$ whenever $\mu(E) < \delta_0$.

Taking $\delta = \min \{\delta_0, \delta_1\}$, whenever $\mu(E) < \delta$, we have $|\int_E f_n d\mu| < \epsilon$ for all $n \in \mathbb{N}$, that is, $\{f_n \}_{n \in \mathbb{N}}$ is uniformly integrable.

  • $\begingroup$ Why should $\|f_n-f_m\|_1$ be small for fixed $n$ and large $m$? $\endgroup$ Nov 11, 2017 at 23:27
  • $\begingroup$ @FriedrichPhilipp I was thinking of the Cauchy property as I wrote it but as soon as I saw your response, I realized the issue. Thanks. $\endgroup$ Nov 11, 2017 at 23:32
  • $\begingroup$ Use that each $f_n$ is uniformly integrable. Given $\epsilon$, divide the sequence into two parts. One that is close to $f$ and... well... the rest. ;-) $\endgroup$ Nov 11, 2017 at 23:35
  • $\begingroup$ You have to show that a finite set is uniformly integrable. This is clear since a single $L^1$ function is uniformly integrable, but you need to show that. Also there's a lot of waste here: Instead of starting with $|f_n|\le|f_n-f_m|+|f_m-f|+|f|$ you could just use $|f_n|\le|f_n-f|+|f|$. $\endgroup$ Nov 12, 2017 at 0:07
  • $\begingroup$ @FriedrichPhilipp I have updated my post with a new proof. This is probably what you had in mind. $\endgroup$ Nov 12, 2017 at 0:07

1 Answer 1


Let $\{f_n\}\rightarrow f\in L^1(\mu)$ and $\epsilon > 0$. Choose $N\in\mathbb{N}$ such that $$\int |f_n - f|d\mu < \frac{\epsilon}{2}$$ for all $n\in\mathbb{N}$ with $n> N$.

First, note that, since $\{f_n\}_\{n=1\}^N$ is uniformly integrable by part a.) there exists a $\delta_1 > 0$ such that, for all $n \leq N$, $$\left|\int_E f_n d\mu\right| < \epsilon$$ whenever $E\in M$ and $\mu(E) < \delta_1$.

Now, since $\{f\}$ is uniformly integrable by part a.) there exists a $\delta_2 > 0$ such that $$\left|\int_E f d\mu\right| < \frac{\epsilon}{2}$$ whenever $E\in M$ and $\mu(E) < \delta_2$. Then, for all $n>N$, we have \begin{align*} \left|\int_E f_n d\mu\right| &= \left|\int_E (f_n - f) + \int_E f d\mu\right|\\ &\leq \left|\int_E (f_n - f)d\mu\right| + \left|\int_E f d\mu\right|\\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2}\\ &= \epsilon \end{align*} whenever $E\in M$ and $\mu(E) < \delta_2$.

So, taking $\delta=\min\{\delta_1, \delta_2\}$, we have that, for all $n$, $$\left|\int_E f_n d\mu\right| <\epsilon$$ whenever $E\in M$ and $\mu(E) < \delta$. Thus $\{f_n\}$ is uniformly integrable.


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