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How can show that that $\lim_\limits{k \to \infty} \sup\limits_{i \in J} E[(|Y_i|-k)^+]=0$ implies uniform integrability of the set of r.v's $(Y_i)_{i \in J}$

I have spent quite some time ,unsuccessfully trying to show that the above statement implies

$$\lim_{k \to \infty} \sup_{i \in J} E[|Y_i| \mathbb{1}_{\{|Y_i|\geq k\}}]=0.$$

It is clear to me that

$E[|Y_i| \mathbb{1}_{\{|Y_i|\geq k\}}] \geq E[(|Y_i|-k)^+]$ so applying sup and lim to both sides of the inequality changes nothing but since

$$\lim_{k \to \infty} \sup_{i \in J} E[(|Y_i|-k)^+]=0,$$

$\lim\limits_{k \to \infty} \sup\limits_{i \in J}E[|Y_i| \mathbb{1}_{\{|Y_i|\geq k\}}] \geq 0$ is not necessarily $0$ I tried several other things but I was unable to prove it. Can someone help me?

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    $\begingroup$ Hint: For every nonnegative random variable $X$ and every nonnegative $k$, $$E(X;X\geqslant 2k)\leqslant2E(X-k;X\geqslant 2k)\leqslant2E(X-k;X\geqslant k)=2E((X-k)^+).$$ $\endgroup$ – Did Oct 17 '15 at 18:56
  • $\begingroup$ So I can say $\lim_{k \to \infty} \sup_{i \in J}E(|Y_i| ;|Y_1| \geq 2k) \leq \lim_{k \to \infty} \sup_{i \in J} E((|Y_i|-k)^{+})=0$ and the fact that we have $2k $ instead of $k$ makes no difference as $k \to \infty$ Right? And since the first expectation on is always positive we have the result. Does this make sense? Thank you so much did $\endgroup$ – user3503589 Oct 17 '15 at 21:32

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