3
$\begingroup$

The term "uniformly integrable" sounds (to a layman like me) to be stronger than integrable. Just like how uniformly convergent is stronger than simply being convergent.

However, from the definition of uniformly integrable, I can't seem to see how if a family of measurable functions is uniformly integrable, then each function is integrable ($\int f_n <\infty$). Thanks for any help! I am sure it is quite trivial but I can't see it immediately.

I post the definition of uniformly integrable as stated in my lecture notes: A family $\mathcal{H}$ of measurable functions on $E$ is said to be uniformly integrable over $E$ provided for each $\epsilon >0$, there is a $\delta>0$ such that for all $f\in\mathcal{H}$, if $A\subseteq E$ is measurable with $m(A)<\delta$, then $\int_A |f|<\epsilon$.

$\endgroup$
5
$\begingroup$

Uniform integrability of a family $\mathcal{H}$ doesn't imply integrability of its members.

If we have a measure such that there is a $c > 0$ so that for each measurable set $A$ we have either $\mu(A) = 0$ or $\mu(A) \geqslant c$, then any family of measurable functions is uniformly integrable. Take $0 < \delta < c$, regardless of $\varepsilon$.

And whatever the measure is, the family $\mathcal{H} = \{ 1\}$ is uniformly integrable (take $\delta = \varepsilon$), but the constant $1$ is only integrable if the whole space has finite measure.

But if the space has finite measure and the measure is atomless, or $\pm\infty$ are excluded as values, then uniform integrability of a family $\mathcal{H}$ implies integrability of its members.

If the measure is atomless, for every $\delta > 0$ you can cover the entire space with finitely many (disjoint) measurable sets $A_1,\dotsc,A_n$ of measure $< \delta$, and hence

$$\int_X \lvert f\rvert \,dm = \int_{A_1} \lvert f\rvert\,dm + \dotsc + \int_{A_n} \lvert f\rvert\,dm < n\cdot \varepsilon < +\infty.$$

If $\pm\infty$ are excluded as values(1), then for every $f\in \mathcal{H}$ we have

$$\bigcap_{n\in\mathbb{N}} \{ x : \lvert f(x)\rvert > n\} = \varnothing,$$

and since $m(X) < +\infty$, there is an $n$ with $m(\{ x : \lvert f(x)\rvert > n\}) < \delta$, so

$$\int_X \lvert f\rvert\,dm = \int_{\{\lvert f\rvert \leqslant n\}} \lvert f\rvert\,dm + \int_{\{ \lvert f\rvert > n\}} \lvert f\rvert\,dm \leqslant n\cdot m(X) + \varepsilon < +\infty.$$

If the measure has atoms and $\pm\infty$ are allowed as values, then an $f\in \mathcal{H}$ could have the value $+\infty$ on an atom, that would not interfere with uniform integrability but make $f$ non-integrable.

(1) Thanks to PhoemueX for the argument.

$\endgroup$
  • 2
    $\begingroup$ +1. Maybe it is useful to note that for finite measure spaces, uniform integrability does imply integrability. $\endgroup$ – PhoemueX Oct 19 '15 at 16:13
  • 1
    $\begingroup$ If you exclude the values $\pm \infty $, you don't need to assume that the measure is atomless (as long as it is finite). Simply note that $M_n =\{x \mid |f (x)|>n\} $ is decreasing with $\bigcap M_n =\emptyset $, so that $\mu (M_n )\to 0$ (here we use that $\mu $ is finite). In particular, $\int_{M_n} |f|d\mu <\infty $ for $n $ large by uniform integrability. But $\int_{M_n^c}|f|d\mu <\infty $ is clear (since $\mu $ us finite). $\endgroup$ – PhoemueX Oct 19 '15 at 20:24
  • 1
    $\begingroup$ @PhoemueX Ah. I had thought of that situation, but took it out again because I couldn't prove that a measurable function must be essentially constant on an atom in my head. I didn't think of the $\bigcap M_n = \varnothing$ route to show integrability in the presence of atoms. Thanks again. $\endgroup$ – Daniel Fischer Oct 19 '15 at 20:35
  • 1
    $\begingroup$ @DanielFischer thanks! Something that is bugging me is how do we ensure that we can "cover the entire space with finitely many disjoint measurable sets $A_i$ with measure $<\delta$"? For instance if we consider the counting measure, each nonempty set has at least measure 1, hence no nonempty set has measure less than, say, half? $\endgroup$ – yoyostein Oct 20 '15 at 1:18
  • 2
    $\begingroup$ @yoyostein The counting measure isn't atomless (it's purely atomic). For an atomless measure, Sierpiński has shown that it attains all values between $0$ and $m(X)$, so with $c = m(X)/n$, there is a measurable subset $A_1$ of measure $c$. Then $m(X\setminus A_1)=(n-1)c$, so (if $n>1$) there is an $A_2\subset X\setminus A_1$ with $m(A_2)=c$. Iterate until you have $n$ parts. If we have a measure with atoms, that of course doesn't generally work. You can then split into finitely many atoms of measure $\ge\delta$ and parts with $m(A)<\delta$. $\endgroup$ – Daniel Fischer Oct 20 '15 at 8:36

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.