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 $\int_A |f|<\epsilon$.
 A: 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.
