Show that this definition of uniform integrability implies the other one Assume $\mathcal{F}$ is set of integrable functions which satifies
$$\forall \varepsilon >0 \exists g_\varepsilon \forall f\in\mathcal{F}: ~ \int\limits_{[\vert f\vert>g_\varepsilon]}\vert f \vert<\varepsilon$$
where $g_\varepsilon$ is an integrable function. We have defined uniform integrability that way and I want to show that this definition implies the following definition of uniform integrability.
So I want to prove now that with the assumption above the following assertions hold for $\mathcal{F}$:


*

*$\sup\{\Vert f\Vert_1 : f\in\mathcal{F}\}<\infty$

*$\lim\limits_{M\rightarrow\infty} \sup\limits_{f\in\mathcal{F}}\int\limits_{[\vert f\vert>M]\cup[\vert f\vert<\frac{1}{M}]}\vert f \vert=0$
I have managed to show the first assertion by assuming that it doesn't hold and concluded a contradiction. I'm stuck with the second one, I also to do a proof by contradiction by assuming that there is at least one $f$ such that $\lim\limits_{M\rightarrow\infty} \int\limits_{[\vert f\vert>M]\cup[\vert f\vert<\frac{1}{M}]}\vert f \vert\neq0$ but I'm not sure how to conclude that there is no integrable function $g_\varepsilon$.
Can someone help me with this exercise?
 A: If there was only one function in $\mathcal F$, then we could apply the dominated convergence theorem. Here, the uniform integrability shows that we are actually not too far away from this situation.
Observe that 
$$
\int\limits_{[\vert f\vert>M]\cup[\vert f\vert<\frac{1}{M}]}\vert f \vert= 
\int\limits_{[\vert f\vert>M]\cup[\vert f\vert<\frac{1}{M}]}\vert f \vert\mathbf 1_{\{\lvert f\rvert\leqslant g_\varepsilon\}}+\int\limits_{[\vert f\vert>M]\cup[\vert f\vert<\frac{1}{M}]}\vert f \vert\mathbf 1_{\{\lvert f\rvert\gt g_\varepsilon\}};
$$
the second term is smaller than $\int\vert f \vert\mathbf 1_{\{\lvert f\rvert\gt g_\varepsilon\}}\lt\varepsilon$. 
The first term can be bounded as follows:
$$
\int\limits_{[\vert f\vert>M]\cup[\vert f\vert<\frac{1}{M}]}\vert f \vert\mathbf 1_{\{\lvert f\rvert\leqslant g_\varepsilon\}}\leqslant \int\limits_{[\vert f\vert>M] }\vert f \vert\mathbf 1_{\{\lvert f\rvert\leqslant g_\varepsilon\}}
+\int\limits_{ [\vert f\vert<\frac{1}{M}]}\vert f \vert\mathbf 1_{\{\lvert f\rvert\leqslant g_\varepsilon\}}\\ 
\leqslant \int\limits_{[g_\varepsilon>M] }\vert f \vert\mathbf 1_{\{\lvert f\rvert\leqslant g_\varepsilon\}}
+\int\limits_{ }\vert f \vert\mathbf 1_{\{\lvert f\rvert\leqslant \min\left\{g_\varepsilon,\frac 1M\right\}\}}\\
\leqslant  \int\limits_{[g_\varepsilon>M] }g_\varepsilon 
+\int \min\left\{g_\varepsilon,\frac 1M\right\}.$$
  We get in total 
$$
\sup_{f\in\mathcal F}\int\limits_{[\vert f\vert>M]\cup[\vert f\vert<\frac{1}{M}]}\vert f \vert\leqslant \varepsilon+\int\limits_{[g_\varepsilon>M] }g_\varepsilon +\int\limits_{[g_\varepsilon\leqslant \frac 1M] }g_\varepsilon  ,
+\int \min\left\{g_\varepsilon,\frac 1M\right\}.
$$
By the dominated convergence theorem, 
$$
\limsup_{M\to +\infty}\sup_{f\in\mathcal F}\int\limits_{[\vert f\vert>M]\cup[\vert f\vert<\frac{1}{M}]}\vert f \vert\leqslant \varepsilon.
$$
