Let $(\Omega,\mathcal{A},\mu)$ be a measurable space and $\mathcal{F}$ a set of measurable functions. Show: If $\mu(\Omega)<\infty$, $\mathcal{F}$ is uniformly integrable exactly then, when for any $\varepsilon > 0$ there exists a constant $a_{\varepsilon}>0$ so that $$ \sup_{f\in\mathcal{F}}\int 1_{\lvert f\rvert\geq a_{\varepsilon}}\lvert f\rvert\, d\mu<\varepsilon. $$


For the direction "$\Leftarrow$" my idea is to use Uniform integrability of a set of measurable functions (show an equivalence), because:

Consider any $\varepsilon > 0$ and any $f\in\mathcal{F}$, then $$ (\lvert f\rvert - a_{\varepsilon})^+\leq 1_{\lvert f\rvert\geq a_{\varepsilon}}\lvert f\rvert $$ and so $$ \int (\lvert f\rvert-a_{\varepsilon})^+\, d\mu\leq\int 1_{\lvert f\rvert\geq a_{\varepsilon}}\lvert f\rvert\, d\mu\leq\sup_{f\in\mathcal{F}}\int 1_{\lvert f\rvert\geq a_{\varepsilon}}\lvert f\rvert\, d\mu<\varepsilon, $$ and therefore $$ \sup_{f\in\mathcal{F}}\int (\lvert f\rvert-a_{\varepsilon})^+\, d\mu<\varepsilon $$ which means (relating to the given link), that $\mathcal{F}$ is uniformly integrable, because $a_{\varepsilon}$ is the required non-negative, integrable function.

For the other direction my idea is the following:

Let $\mathcal{F}$ be uniformly integrable. Consider any $\varepsilon >0$. Then it exists a non-negative, integrable function $h$ so that $$ \sup_{f\in\mathcal{F}}\int 1_{\lvert f\rvert\geq h}\lvert f\rvert\, d\mu<\varepsilon/2. $$ Now choose $a_{\varepsilon}$, so that $$ \int 1_{h\geq a_{\varepsilon}}h\, d\mu<\varepsilon/2. $$ To my opinion it is $$ 1_{\lvert f\rvert\geq a_{\varepsilon}}\lvert f\rvert\leq 1_{\lvert f\rvert\geq h}\lvert f\rvert+1_{h\geq a_{\varepsilon}}h, $$ so it is $$ \int 1_{\lvert f\rvert\geq a_{\varepsilon}}\lvert f\rvert\, d\mu\leq\int 1_{\lvert f\rvert\geq h}\lvert f\rvert\, d\mu+\int 1_{h\geq a_{\varepsilon}}h\, d\mu\\\leq \sup_{f\in\mathcal{F}}\int 1_{\lvert f\rvert\geq h}\lvert f\rvert\, d\mu+\int 1_{h\geq a_{\varepsilon}}h\, d\mu<\varepsilon/2 + \varepsilon/2=\varepsilon. $$

But where do I have to use that $\mu(\Omega)<\infty$?

To my opinion it is necessary for the direction "$\Leftarrow$", because $a_{\varepsilon}$ is only integrable if $\mu(\Omega)<\infty$, because then $$ \int\lvert a_{\varepsilon}\rvert\, d\mu=a_{\varepsilon}\mu(\Omega)<\infty. $$ Are there more points in the proof, where I need $\mu(\Omega)<\infty$?

By the way: What do you think about my proof?

Sincerely yours,


  • 1
    $\begingroup$ Since this is often taken as the definition of uniform integrability, you might want to add the definition you are considering. Upvoters: surely you know how to evade the paradox? $\endgroup$ – Did Dec 24 '13 at 13:00
  • $\begingroup$ Hello, I use the definition of uniformly integrable as given in the link. There I give our definition. But if you say that I should write it down here, I do. $\endgroup$ – math12 Dec 24 '13 at 13:03
  • $\begingroup$ Since this is a highly idiosyncratic definition (actually nobody considers it as such), you should definitely include it. (Note that you now consider as a definition a property you asked to prove 20 hours ago.) $\endgroup$ – Did Dec 24 '13 at 14:01
  • $\begingroup$ Both equivalences (the linked and the one here) are to show. I use the equivalence in the link, assuming that it is proved. $\endgroup$ – math12 Dec 24 '13 at 14:08

Since the measure of the space is finite, the choice $h$ constant is allowed provided the constant is large enough. I don't think there are other points were it is used.

The proof is correct I think, but I have a remark: avoid in a proof "in my opinion", here we can give more details. For example, write $$\chi_{|f|\gt a_\varepsilon}|f|\leqslant |f|\chi_{|f|\gt h}+|f|\chi_{a_\varepsilon\lt |f|\leqslant h}\leqslant |f|\chi_{|f|\gt h}+h\chi_{a_\varepsilon \leqslant h}.$$

  • 1
    $\begingroup$ What do you mean with "large enough"? My argument above for the direction "$\Leftarrow$" is that for given $\varepsilon > 0$ I can use $a_{\varepsilon}$ as the function $h$ in the definition. But because $h$ has to be integrable, it has to be $\mu(\Omega)<\infty$, because only then it is $\int\lvert a_{\varepsilon}\rvert\, d\mu<\infty$. -- Do you mean that? $\endgroup$ – math12 Dec 24 '13 at 13:26
  • $\begingroup$ Yes, that's what I mean. $\endgroup$ – Davide Giraudo Dec 24 '13 at 13:27
  • $\begingroup$ Great, thank you! Do I need $\mu(\Omega)<\infty$ at other points in the proof? By the way: What do you think of my proof? $\endgroup$ – math12 Dec 24 '13 at 13:28
  • $\begingroup$ I've edited. Is is better? $\endgroup$ – Davide Giraudo Dec 24 '13 at 22:31
  • $\begingroup$ I thank you very much! Really great help!! $\endgroup$ – math12 Dec 24 '13 at 23:41

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