# Uniformly integrable implies integrable?

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 $\int_A |f|<\epsilon$.

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.

• +1. Maybe it is useful to note that for finite measure spaces, uniform integrability does imply integrability. – PhoemueX Oct 19 '15 at 16:13
• 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). – PhoemueX Oct 19 '15 at 20:24
• @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. – Daniel Fischer Oct 19 '15 at 20:35
• @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? – yoyostein Oct 20 '15 at 1:18
• @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 $m(A)<\delta$. – Daniel Fischer Oct 20 '15 at 8:36