5
$\begingroup$

Let $(X, \mathcal{A}, \mu)$ be a measure space. A family of measurable functions $\{f_n\}$ is uniformly integrable if given $\epsilon$ there exists $M$ such that$$\int_{\{x : |f_n(x)| > M\}} |f_n(x)|\,d\mu < \epsilon$$for each $n$. The sequence is uniformly absolutely continuous if given $\epsilon$ there exists $\delta$ such that$$\left|\int_A f_n\,d\mu\right| < \epsilon$$for each $n$ if $\mu(A) < \delta$.

Suppose $\mu$ is a finite measure. How do I see 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?

$\endgroup$
3
$\begingroup$

Assume that $\{f_n\}$ is uniformly absolutely integrable. Let $\varepsilon > 0$. Now \begin{align} \int_X |f_n| = \int_{X \cap \{f_n \geq M_\varepsilon\}} |f_n| + \int_{X \cap \{f_n < M_\varepsilon\}} |f_n| \leq \varepsilon + M_\varepsilon \mu(X) \end{align} for all $n$. Thus the supremum over $n$ is finite.

To get uniform absolute continuity, notice that \begin{align} \left| \int_A f_n \right| & \leq \int_{A \cap \{f_n \geq M_\varepsilon\}} |f_n| + \int_{A \cap \{f_n < M_\varepsilon\}} |f_n| \\ & \leq \varepsilon + M_\varepsilon \mu(A) \end{align} for all $n$. Now choose $\delta < \varepsilon/M_\varepsilon$.

Now assume $\sup_n \|f_n\|_1 < \infty$ and the uniform abs. continuity. Let $\varepsilon > 0$ and let $\delta > 0$ be such that $\mu(A) < \delta$ implies $ |\int_A f_n| < \varepsilon$ for all $n$. Since $\int|f_n| < \infty$, we have \begin{align} \lim_{M \to \infty} \mu \{ |f_n| > M \} = 0\,. \end{align} Thus we may choose $M_n$ so large that $\mu\{ |f_n| > M_n \} < \delta$. Now \begin{align} \int_{|f_n| > M_n } |f_n| &= \int_{f_n > M_n} f_n + \int_{f_n < -M_n} (-f_n) \\ &= \left| \int_{f_n > M_n} f_n \right| + \left| \int_{f_n < -M_n} f_n \right| \\ &< \varepsilon + \varepsilon \end{align} for all $n$ since the sets over which we integrate have measure less than $\delta$

$\endgroup$
  • 1
    $\begingroup$ The reverse direction is incorrect. To establish uniform integrability, $M$ cannot depend on $n$. To prove it, first fix $\epsilon>0$ and choose $\delta>0$ be such that $\mu(A)<\delta$ implies $\vert \int_A f_n\vert <\epsilon$. Let $N:= \max\{0, \sup_n \|f_n\|_{L^1}\}$ and choose $M> \frac{N}{\delta}$. By Markov's inequality, $$\mu\{\vert f_n\vert>M\}\le M^{-1} \int_X \vert f_n\vert \le M^{-1}N<\delta.$$ Thus, by the uniform absolutely continuity condition, $$ \int_{\{\vert f_n\vert>M\}} \vert f_n\vert <\epsilon,$$ so the proof is now complete. $\endgroup$ – Satana Feb 3 '18 at 19:13
2
$\begingroup$

Here I show the forward direction. Once I see the reverse direction, I will edit.

Given $\varepsilon>0$, choose $M>0$ so that $$ \int_{\{x : |f_n(x)| > M\}} |f_n(x)|\,d\mu < \epsilon $$ for all $n$. Then we have $$ \int|f_n|\ d\mu=\int_{\{|f_n|\leq M\}}|f_n|\ d\mu+ \int_{\{|f_n|> M\}}|f_n|\ d\mu\leq M\mu(X)+\varepsilon<\infty.$$ Since $n$ was arbitrary, we have $$\sup_n\int|f_n|\ d\mu<\infty. $$ Now pick $\delta>0$ so that $\delta<\varepsilon/M$. Then for measurable $A$ with $\mu(A)<\delta$, we have \begin{align*} \left|\int_A f_n\ d\mu\right|&\leq \left|\int_{A\cap\{|f_n|\leq M\}} f_n\ d\mu\right| +\left|\int_{A\cap\{|f_n|> M\}} f_n\ d\mu\right| \\ &\leq \int_{A\cap\{|f_n|\leq M\}} |f_n|\ d\mu + \int_{A\cap\{|f_n|> M\}} |f_n|\ d\mu \\ &<M\delta +\varepsilon <2\varepsilon. \end{align*} By rescaling, we see that $\{f_n\}$ is uniformly absolutely continuous.

$\endgroup$
  • $\begingroup$ Doesn't this presume that the space has finite measure? That's not always the case. $\endgroup$ – rem Mar 11 at 6:35
  • $\begingroup$ @rem The question specifically asks for the finite measure case. $\endgroup$ – Aweygan Mar 11 at 6:37
  • $\begingroup$ Oh of course my bad! $\endgroup$ – rem Mar 11 at 6:41
  • $\begingroup$ No problem, it happens. $\endgroup$ – Aweygan Mar 11 at 6:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.