Let $(\Omega,\mathcal{A},\mu$ be a measure space.

A set $\mathcal{F}$ of measurable, numerical functions is called equi-integrable if for any $\varepsilon > 0$ it exists a nonnegative, integrable function $h$ such that $$ \sup_{f\in\mathcal{F}}\int 1_{\left\{\lvert f\rvert\geq h\right\}}\lvert f\rvert\, d\mu < \varepsilon. $$

Now the task is to show the following:

$\mathcal{F}\subset\mathcal{L}_{\mu}^1$ finite $\implies \mathcal{F}$ equi-integrable.

My idea to prove that is:

Let $\mathcal{F}=\left\{f_1,\ldots,f_n\right\}$ with $f_i\in\mathcal{L}_{\mu}^1$ for $i=1,\ldots,n$. This means that $0\leq\lvert f_i\rvert=m_i<\infty$ a.s. for $i=1,\ldots.n$. Choose $$ h:=\max_{i\in\left\{1,\ldots,n\right\}}\left\{m_i\right\}. $$ Then for any $\varepsilon > 0$ and any $f_i\in\mathcal{F}$ it is $$ \int 1_{\left\{\lvert f_i\rvert\geq h\right\}}\lvert f_i\rvert\, d\mu=0<\varepsilon $$ and because this was not dependent on $i$, it is for any $\varepsilon >0$ $$ \sup_{f\in\mathcal{F}}\int 1_{\left\{\lvert f\rvert\geq h\right\}}\lvert f\rvert\, d\mu=0<\infty. $$

So $\mathcal{F}$ is equi-integrable.

  • $\begingroup$ Got something from the answer below? $\endgroup$
    – Did
    Mar 4, 2015 at 17:24
  • $\begingroup$ Apparently not. Anyway, bad manners. $\endgroup$
    – Did
    Sep 9, 2016 at 10:45

1 Answer 1


Try $h=2\sum\limits_{f\in\mathcal F}|f|$. Let $A=\bigcap\limits_{f\in\mathcal F}\{f=0\}$. Then, for every $f$ in $\mathcal F$, $\{|f|\geqslant h\}=A$ and $\int\limits_A|f|\mathrm d\mu=0$.

Additionally, note that the argument based on the fact that for every $f$ in $\mathcal L^1_\mu$ there would exist some finite constant $c$ such that $|f|\leqslant c$ almost surely, is wrong (try $f(x)=1/\sqrt{x}$ on $\Omega=(0,1)$ with $\mathcal A$ the Borel sigma-algebra and $\mu$ the Lebesgue measure).

  • $\begingroup$ This is only the case for measurable, non-negative functions, right? $\endgroup$
    – mathfemi
    Feb 11, 2014 at 14:45
  • $\begingroup$ Nonnegative: no. Measurable: yes, but anyway, $\mathcal F$ is made of measurable functions, isn't it? $\endgroup$
    – Did
    Feb 11, 2014 at 14:49
  • $\begingroup$ In our reading we had: Let $f$ be a non-negative, measurable, numerical function then: $\int f\, d\mu<\infty\implies f<\infty$ a.s.. $\endgroup$
    – mathfemi
    Feb 11, 2014 at 14:51
  • $\begingroup$ Sure, but being almost surely finite and being bounded are not the same condition. Did you read the example in my post? $\endgroup$
    – Did
    Feb 11, 2014 at 14:53
  • $\begingroup$ Sorry, I do not chat. $\endgroup$
    – Did
    Feb 11, 2014 at 14:54

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