In his nice answer about uniform integrability, Did did show that condition (C) below is equivalent to (C1) and (C2) together.
(C) For every $\varepsilon\gt0$, there exists a finite $c$ such that, for every $X$ in $\mathcal H$, $\mathrm E(|X|:|X|\geqslant c)\leqslant\varepsilon$.
(C1) There exists a finite $C$ such that, for every $X$ in $\mathcal H$, $\mathrm E(|X|)\leqslant C$.
(C2) For every $\varepsilon\gt0$ there exists $\delta\gt0$ such that, for every measurable $A$ such that $\mathrm P(A)\leqslant\delta$ and every $X$ in $\mathcal H$, $\mathrm E(|X|:A)\leqslant\varepsilon$.
I am wondering why (C2) does not imply (C), alone. Would anyone be so nice as to give me a counterexample? I mean, is there a family of random variables $\mathcal H$ such that (C2) holds but (C) (or (C1)) does not hold?
Following Did's comment to my question...
Just take the probability space $(\Omega, \sigma, P)$, where $\Omega = \{a\}$ is a singleton. Or just take a Dirac delta $P = \delta_a$ over any nonempty $\Omega$. In this case, any set of random variables will satisfy (C2).
In particular, $X_n = n$ will satisfy (C2), but not (C1).
Of course, this can easily be adapted to any $P$ with an "atom".
Intuition failed to me because I was under the impression that I could always break up the spaces into "small bits". Suppose that (C2) holds and also that for any $\delta > 0$ there are $A_1, \dotsc, A_n \in \sigma$ such that $P(A_j) < \delta$ and $$ \Omega = A_1 \cup \dotsb \cup A_n. $$ Take $\delta$ as in (C2) for $\varepsilon = 1$. Then, for any $X \in \mathcal{H}$, $$ E(|X|) \leq E(|X|; A_1) + \dotsb + E(|X|; A_n) \leq n. $$
Thank you, Did! :-)