Let $E_M=\{x: \vert \ f_n (x)\ \vert>M\}$. A family of measurable functions $\{f_n\}$ is uniformly integrable if given $\epsilon >0$, there exists $M$ such that $$\int_{E_M} |f_n| \ d\mu<\epsilon$$ for each $n$.

The family is uniformly absolutely continuous if given $\epsilon >0$, there exists $\delta$ such that $$\bigg|\int_A f_n \ d\mu \bigg|< \epsilon$$ for each $n$ if $\mu(A)<\delta$.

Problem Statement

Suppose $\mu$ is a finite measure. Prove that $\{f_n\}$ is uniformly integrable if and only if $\sup_n \int|f_n| \ d\mu <\infty$ and $\{f_n\}$ is uniformly absolutely continuous.


Proposition 1: If $\mu$ is finite, $$\lim_{M\to\infty} E_M \rightarrow 0.$$ Proof: If $E_M \to K$, where $k>0$, then $\infty=\sum_{i=1}^{\infty} K \leq \mu(X)=\mu(\bigcup_{M=1}^{\infty} E_M)$, since $E_{N+1} \subset E_{N}$. This contradicts $\mu$ being finite.

Next I provide a proof of the problem:

Proof:$(\Rightarrow)$ Suppose that $\{f_n\}$ is uniformly integrable. Then, for any $\epsilon>0$, we may find $M$ so that $$ \bigg|\int_{E_M} f_n \ d\mu \bigg| \leq \int_{E_M} |f_n| \ d\mu <\epsilon.$$ Thus, taking $\delta=\mu(E_M)$ is sufficient. Next, note that \begin{align*} \int |f_n| \ d\mu&=\int_{E_M} |f_n|\ d\mu+\int_{E^c_M} |f_n| \ d\mu\\ &\leq \epsilon + M\mu(E^c_M)\\ &\leq \infty, \end{align*} since $\mu$ is finite. SO $\sup_n \int |f_n| \ d\mu\leq \infty$.

$(\Leftarrow)$ On the other hand, assume $\{f_n\}$ is uniformly absolutely continuous and $\sup_n \int |f_n| \ d\mu<\infty$. Let $E^{'}_M=\{x:f_n(x)>M\}$ and $E^{''}_M=\{x:f_n<-M\}$. Given $\epsilon>0$, use proposition 1 to find $E_M$ with $\mu(E_M)<\delta$ and absolute continuity along with the observation that $\mu(E^{'}_M),\mu(E^{''}_M)\leq \mu(E_M)$ to ensure $\bigg|\int_{E^{'}_M} f_n \ d\mu \bigg|$, and $\bigg|\int_{E^{''}_M} f_n \ d\mu \bigg|$ are both less than $\frac{\epsilon}{2}.$ Then, \begin{align*} \int_{E_M}|f_n| \ d\mu&=\int_{E^{'}_M}f_n \ d\mu+\int_{E^{''}_M}-f_n \ d\mu\\ &\leq \bigg|\int_{E^{'}_M} f_n \ d\mu \bigg|+\bigg|\int_{E^{''}_M} f_n \ d\mu \bigg| \\ &\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}\\ &=\epsilon \end{align*}


Is my proof correct? Suggestions are greatly appreciated.


1 Answer 1


The proof is incorrect. In the fist part of the proof, the $\delta$ you picked does not guarantee that for all measurable sets $A$, $\mu(A) < \delta$ implies $|\int_A f_n\, d\mu| < \epsilon$. The second part of the proof is almost correct, but the condition $\lim\limits_{M\to \infty} \mu(E_M) = 0$ follows from the assumption $\sup_n \int |f_n|\, d\mu < \infty$, not that $\mu$ is finite. Indeed, if $\alpha := \sup_n \int |f_n|\, d\mu < \infty$, then $\mu(E_M) \le \frac{\alpha}{M}$ for all $M > 0$; as a consequence, $\lim\limits_{M\to \infty} \mu(E_M) = 0$. If you make this fix, then the second part of the proof will be correct.

To prove the forward direction, fix $\epsilon > 0$ and choose $M > 0$ such that

$$\sup_n \int_{E_M} |f_n| \, d\mu < \frac{\epsilon}{2}.$$


$$\int |f_n|\, d\mu = \int_{E_M} |f_n|\, d\mu + \int_{E_M^c} |f_n|\, d\mu < \frac{\epsilon}{2} + M\mu(X) < \infty$$

for all $n\in N$.

Set $\delta = \frac{\epsilon}{2M}$. For all measurable sets $A$, $\mu(A) < \delta$ implies

$$\left|\int_A f_n\, d\mu\right| \le \int_A |f_n|\, d\mu = \int_{A\cap E_M} |f_n|\, d\mu + \int_{A\cap E_M^c} |f_n|\, d\mu < \frac{\epsilon}{2} + M\mu(A) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

Since $\epsilon$ was arbitrary, $\{f_n\}$ is uniformly absolutely continuous.

  • $\begingroup$ Thank-you for the response... I see now why my forward implication of incorrect but I am having trouble seeing why my proposition 1 is wrong $\endgroup$
    – illysial
    Jun 19, 2015 at 22:20
  • $\begingroup$ Hi @illysial, the claim of proposition 1, i.e., $\lim\limits_{M\to \infty} E_M = 0$, does not make sense. I assume you mean $\lim\limits_{M\to \infty} \mu(E_M) = 0$. In the proof of the proposition, you wrote the inequality $\sum\limits_{i = 1}^\infty K \le \mu(X)$. That inequality need not hold. $\endgroup$
    – kobe
    Jun 20, 2015 at 1:27
  • $\begingroup$ Suppose we assume the $f_n$ are bounded in $L^1$. Then is the requirement of $\mu$ finite unnecessary? $\endgroup$
    – FShrike
    Feb 24 at 9:25
  • $\begingroup$ @FShrike finiteness of $\mu$ is unnecessary in proving the reverse direction. It is only needed in the forward direction. $\endgroup$
    – kobe
    Feb 25 at 14:26
  • $\begingroup$ @kobe Thank you. I notice on Wikipedia they say that, in general for any possibly infinite space, the two are equivalent (without the boundedness assumption) if you only assume each of the functions are in $L^1$ and the space is atomless $\endgroup$
    – FShrike
    Feb 25 at 14:58

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