We call a set $\mathcal{F}$ of measurable functions uniformly integrable if for any $\varepsilon >0$ it exists a non-negative, integrable function $h$ so that $$ \sup_{f\in\mathcal{F}}\int 1_{\left\{\lvert f\rvert\geq h\right\}}\lvert f\rvert\, d\mu<\varepsilon.~~~~~~(*) $$ Now the task is to show the following:
Let $(\Omega,\mathcal{A},\mu)$ be a measurable space and $\mathcal{F}$ a set of measurable functions. Show, that $\mathcal{F}$ is equi-integrable exactly when for any $\varepsilon >0$ there exists an integrable function $h\geq 0$ so that $$ \sup_{f\in\mathcal{F}}\int (\lvert f\rvert-h)^+\, d\mu\leq\varepsilon. $$
Here is how I would prove it.
"$\Longrightarrow$": Consider any $\varepsilon >0, f\in\mathcal{F}$. It is $$ (\lvert f\rvert -h)^+\leq 1_{\lvert f\rvert\geq h}\lvert f\rvert $$ and therefore $$ \int (\lvert f\rvert-h)^+\, d\mu\leq\int 1_{\lvert f\rvert\geq h}\lvert f\rvert\, d\mu<\varepsilon. $$ Because the right side does not depend on $f$, it is $$ \sup_{f\in\mathcal{F}}\int (\lvert f\rvert-h)^+\, d\mu<\varepsilon. $$
"$\Longleftarrow$": Consider any $\varepsilon > 0$. It exists a $h_{\varepsilon/2}\geq 0$, $h_{\varepsilon/2}$ integrable with $$ \sup_{f\in\mathcal{F}}\int (\lvert f\rvert-h_{\varepsilon/2})^+\, d\mu<\varepsilon/2.~~~~~(+) $$ Define $$ k:=2h_{\varepsilon/2}. $$ For any$f\in\mathcal{F}$ it is with the $\Delta$-inequation $$ \lvert f\rvert\leq\lvert\lvert f\rvert-h_{\varepsilon/2}\rvert+\lvert h_{\varepsilon/2}\rvert=\lvert\lvert f\rvert-h_{\varepsilon/2}\rvert+h_{\varepsilon/2}. $$ Because of $$ (\lvert f\rvert-h_{\varepsilon/2})^+=\lvert f\rvert-h_{\varepsilon/2} $$ for $\lvert f\rvert \geq k$, i.e. especially $\lvert f\rvert\geq h_{\varepsilon/2}$, it is $$ 1_{\lvert f\rvert\geq k}\lvert f\rvert\leq 1_{\lvert f\rvert\geq k}((\lvert f\rvert-h_{\varepsilon/2})^++h_{\varepsilon/2}) $$ and therefore $$ \int 1_{\lvert f\rvert\geq k}\lvert f\rvert\, d\mu\leq\int1_{\lvert f\rvert\geq k}(\lvert f\rvert-h_{\varepsilon/2})^+\, d\mu+\int 1_{\lvert f\rvert\geq k}h_{\varepsilon/2}\, d\mu. $$ It is $$ 1_{\lvert f\rvert\geq k}(\lvert f\rvert-h_{\varepsilon/2})^+\leq (\lvert f\rvert-h_{\varepsilon/2})^+ $$ and because of $(+)$ it follows that $$ \int 1_{\lvert f\rvert\geq k}(\lvert f\rvert-h_{\varepsilon/2})^+\, d\mu<\varepsilon/2 $$ and because of $$ 1_{\lvert f\rvert\geq k}h_{\varepsilon/2}\leq 1_{\lvert f\rvert\geq k}(\lvert f\rvert-h_{\varepsilon/2})^+ $$ it is in the same way $$ \int 1_{\lvert f\rvert\geq k} h_{\varepsilon/2}\, d\mu < \varepsilon /2, $$ what means that $$ \int 1_{\lvert f\rvert\geq k}\lvert f\rvert\, d\mu<\varepsilon. $$ Because the right side does not depend on $f$, it follows that $$ \sup_{f\in\mathcal{F}}\int 1_{\lvert f\rvert\geq k}\lvert f\rvert\, d\mu<\varepsilon, $$ i.e. a function $h$ is found which fullfills the definition, namely $k$.
That's it. Could anybody say me, if my proof is allright?
The proof looks correct. Maybe it can be shortened a little, skipping some estimates like $|f|\leqslant |f-h|+|h|\color{blue}{=|f-h|+h}$ since$h$ is non-negative.