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?


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 $\delta$

  • 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

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.

  • $\begingroup$ 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

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