# If $\mu$ is finite, then $\{f_n\}$ is uniformly integrable iff $\sup_n \int|f_n| d\mu<\infty$ and $\{f_n\}$ is uniformly absolutely continuous.

Background

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.

Attempt

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*}

Question

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

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

Then

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

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.

• 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 Jun 19 '15 at 22:20
• 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.
– kobe
Jun 20 '15 at 1:27