I want to prove:
$E \left[ |X| \right] < \infty \Leftrightarrow \forall \varepsilon > 0 ~\exists \delta > 0 $ such that $E \left[ |X| \mathbb{1}_{A}\right]< \varepsilon ~\forall A \in \mathcal{F}$ with $P(A) < \delta$
I have already shown "$\Rightarrow$" but I'm struggling with "$\Leftarrow$".
An idea of mine would be: Let $ A = \{ |X| \geq n\} $. We have that $ \mathbb{P}(A)= \int \mathbb{1}_{\{ |X| \geq n\}} \mathrm{d} \mathbb{P} \rightarrow 0 $ for $ n \rightarrow \infty $, because $ \{ |X| \geq n\} \rightarrow \emptyset $
This means that $ \exists K \in \mathbb{N}$, such that $\mathbb{P}\left(A\right) < \delta ~\forall n \geq K $. Thus, $ \mathbb{E} \left( |X| \mathbb{1}_A \right) < \varepsilon $. Make use of this set $ A $ and consider $ \mathbb{E}\left( |X| \right)$.
$ \mathbb{E}\left( |X| \right) = \int |X| \mathbb{1}_ {\{ |X| \geq n\}} \mathrm{d}\mathbb{P} + \int |X| \mathbb{1}_{ \{ |X| < n\}} \mathrm{d}\mathbb{P} < \varepsilon +\int |X| \mathbb{1}_{\{ |X| < K\}}\mathrm{d}\mathbb{P} < \varepsilon + K \int \mathbb{1}_{\{ |X| < K\}}\mathrm{d}\mathbb{P} \leq \varepsilon + K $ This shows that $ \mathbb{E}\left( |X| \right) < \infty ~ \forall A \in \mathcal{F}$ mit $ \mathbb{P}\left( A \right) < \delta$
Is this correct?
(First idea, which is wrong: "Assume $E \left[ |X| \right] = \infty$. Let $A \in \mathcal{F}$ with $P(A) < \delta$. Thus, $ E \left[ |X| \mathbb{1}_{A}\right] = E \left[ |X| \right] P(A) = \infty $. Contradiction.")