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Defintion (Uniform Integrability): A family $\mathcal{F}$ of integrable functions is uniformly integrable if $\forall \varepsilon > 0 $ there is a $M_\varepsilon>0$, such that

$\int_{\{|f|>M_\varepsilon\}}|f|\,\mathrm{d}\mu < \varepsilon,\ \forall f\in\mathcal{F}$

Let $(X,\mathcal{A},\mu)$ be a finite measure space. A family of integrable functions $\mathcal{F}$ is uniformly integrable if and only if for all $\varepsilon > 0$ there exists a $\delta>0$ sucht that for all $A\in\mathcal{A}$ we have that

$\mu(A)<\delta\ \Rightarrow\ \int_A|f|\,\mathrm{d}\mu<\varepsilon ,\ \forall f\in \mathcal{F}$

Here is my proof and I would like to know if there are errors and would be thankful for any improvements.

"$\Rightarrow$":

Because of uniform integrability we can choose $\varepsilon/2 >0$ and get a $M_{\varepsilon/2}$ such that for arbitrary $A\in\mathcal{A}$

$\int_A |f|\,\mathrm{d}\mu =\int_{\{|f|>M_{\varepsilon/2}\}\cap A}|f|\,\mathrm{d}\mu + \int_{\{|f|\le M_{\varepsilon/2}\}\cap A}|f|\,\mathrm{d}\mu\le\mu(A) M_{\varepsilon/2} + \int_{\{|f|>M_{\varepsilon/2}\}\cap A}|f|\,\mathrm{d}\mu$

Note that we have $\mu(A)<\infty$ since the measure space is finite. Now we can pick a $\delta:=\frac{\varepsilon}{2}\frac{1}{M_{\varepsilon/2}}$ from which it follows by the previous equation that for all $A$ satisfying $\mu(A)<\delta$

$\int_A |f|\,\mathrm{d}\mu\le \varepsilon/2 + \varepsilon/2\le \varepsilon $

which is what we wanted to show.

"$\Leftarrow$":

We have, since each member in $\mathcal{F}$ is integrable, that

$\mu\left(\{|f|>m\}\right)\rightarrow 0$ for $m\rightarrow\infty$ and arbitrary $f\in\mathcal{F}$. Any suggestions on how to prove this statement rigorously?

This is equivalent to (just use the definition of a limit and exchange $\delta$ for $\epsilon$)

$\forall \delta >0\ \exists M_\varepsilon\in \mathbb{N}\ \forall n\ge M_\varepsilon\colon \mu\left(\{|f|>n\}\right)\le |\mu\left(\{|f|>n\}\right)|<\delta$

which proves this direction.

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  • $\begingroup$ The wikipedia page shows the thing you are trying to prove as the definition of uniform integrability (which makes me wonder what alternative definition you are assuming): en.wikipedia.org/wiki/Uniform_integrability $\endgroup$
    – Michael
    Dec 18 '19 at 16:38
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    $\begingroup$ In your $\implies$ part, that looks good as long as you clarify that (i) when you choose $\delta = \frac{\epsilon}{2M}$, you are considering sets $A$ such that $\mu(A)<\delta$; (ii) Your original choice of $M_{\epsilon/2}$ is to satisfy the given property of the definition of UI. [I cannot follow your proof of the $\impliedby$ part.] $\endgroup$
    – Michael
    Dec 18 '19 at 17:07
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    $\begingroup$ As an alternative approach to the $\impliedby$ part: If you can show $\sup_f \int |f|<\infty$ then you can use a version of the "Markov inequality" $\int |f| \geq M\mu(\{x : |f(x)|>M\})$ $\endgroup$
    – Michael
    Dec 18 '19 at 17:24
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    $\begingroup$ This post: math.stackexchange.com/questions/1862444/… gives the same definition of uniformly integrability but the equivalent condition contains an additional statement. Maybe you wanna have a look there (be careful: the first answer also gives a false proof for the reverse direction). $\endgroup$
    – Falrach
    Dec 18 '19 at 17:48
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    $\begingroup$ @Falrach This is interesting, since this additional statement allows me to conclude from one particular function to every function in the second part of the proof. I think it is even necessary to have this additional statement. $\endgroup$ Dec 18 '19 at 18:12
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The Falrach link identifies an additional requirement $\sup_{f \in \mathcal{F}} \int |f|d\mu<\infty$ for the posted condition to imply UI. This additional requirement immediately enables the Markov inequality approach from my hint in comments above. Here is a simple counter-example that shows what can go wrong without that additional requirement:

Counter-example:

Define $X=0$ (a 1-element set) with $\mu(X)=1$. For $n \in \{1, 2, 3, ...\}$ define $f_n:X\rightarrow\mathbb{R}$ by $f_n(x)=n$. Clearly $\int_X |f_n|d\mu = n$ for all $n \in \{1, 2, 3, ...\}$ and so $\{f_n\}_{n=1}^{\infty}$ is not UI. But the functions $\{f_n\}_{n=1}^{\infty}$ satisfy the condition of the above post trivially: For all $\epsilon>0$ we can choose $\delta=1/2$ and indeed for any set $A \subseteq X$ that satisfies $\mu(A)<\delta$ we immediately have $\int_A |f_n|d\mu<\epsilon$. This is because the only subset of $X$ with measure less than $1/2$ is the empty set!

So the additional requirement $\sup_{f \in \mathcal{F}} \int |f|d\mu$ is quite needed in general.

A "Covering Property" that implies the additional requirement:

Suppose $\mathcal{F}$ is a family of integrable functions $f:X\rightarrow\mathbb{R}$ such that for all $\epsilon>0$ there is a $\delta>0$ such that $A \subseteq X$ with $\mu(A)<\delta$ implies $\int_A |f|d\mu < \epsilon$ for all $f \in \mathcal{F}$. Now fix $\epsilon=1$ and choose the corresponding $\delta$ so that $\mu(A)<\delta$ implies $\int_A |f|d\mu < 1$ for all $f \in \mathcal{F}$.

If there exists a finite sequence of sets $\{A_1, ...,A_m\}$ (for some positive integer $m$) such that $\cup_{i=1}^m A_i = X$, $A_i\subseteq X$ for all $i \in \{1, ...,m\}$, and $\mu(A_i)<\delta$ for all $i \in \{1, ..., m\}$ then for all $f \in \mathcal{F}$: $$ \int_X |f|d\mu \leq \sum_{i=1}^m \int_{A_i}|f|d\mu \leq m$$ So the additional requirement always holds in this case.

Such a covering by finite sets is always possible when $X$ is a compact subset of $\mathbb{R}^k$ for some positive integer $k$ (and when we use the standard measure for $\mathbb{R}^k$).

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  • $\begingroup$ So I think the wikipedia link from my very first comment is buggy: It gives an incorrect definition of UI. $\endgroup$
    – Michael
    Dec 19 '19 at 1:13
  • $\begingroup$ You should not name it after my name. $\endgroup$
    – Falrach
    Dec 19 '19 at 7:02
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    $\begingroup$ @Michael The property of finding a partition in a measure space with the properties you stated is satisfied in every non-atomic and finite measure space, by the way! This makes it even more interesting, that you can conclude from this that the integrals are bounded! Where did you read of this property? How did you know that the additional requirement holds in non-atomic finite measure spaces? $\endgroup$ Dec 19 '19 at 9:44
  • $\begingroup$ @Falrach : Sorry. I have now changed "Falrach requirement" to "additional requirement" in my answer, although I still point out the "Falrach link" to connect to the history of comments. $\endgroup$
    – Michael
    Dec 19 '19 at 12:40
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    $\begingroup$ I see. I had never heard of the Sierpinksi theorem before, so now I have learned about that. Thanks for sharing the interesting problem and comments! $\endgroup$
    – Michael
    Dec 19 '19 at 13:12

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