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.


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 '15 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 '15 at 1:27

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