The remark of interest is given in my measure-theory script but without a proof and I don't know how to proof it. I am using the following definition. Let in the following $(\Omega,\mathfrak{A},\mu)$ be a measure space.
$\newcommand\diff{\mathop{}\!\mathrm{d}}$ $\newcommand\integrable{\bar{\mathfrak{L}}^1(\mu)}$ $\newcommand\integrableP{\bar{\mathfrak{L}}^1_+(\mu)}$
Def.: (1) A familiy of generalised integrable functions (generalised means that the functions can have values in $\bar{\mathbb{R}}=\mathbb{R}\cup\{\pm\infty\}$) $\mathfrak{F}\subseteq\integrable$ is called $\mu$-uniformly integrable iff for all $\epsilon>0$ there exists a generalised nonnegative integrable function $h\in\integrableP$ such that $$\sup_{f\in\mathfrak{F}}{\int_\mu{|f|\chi_{\{|f|\geq h\}}\diff\omega}}<\epsilon.$$
Remark 1 The above definition (1) is equivalent to definition (2): For all $\epsilon>0$ we have $h'\in\integrableP$ so that $$\sup_{f\in\mathfrak{F}}{\int_\mu{(|f|-h')^+\diff\omega}}<\epsilon.$$
Remark 1 is fine for now. I want to show that any familiy of generalised integrable functions satisfying definition (1) also satisfies definition (3). Of course we can use remark 1.
Remark 2 If $\mu$ is finite then the above definition (1) is equivalent to definition (3) $$\lim_{n\to\infty}\sup_{f\in\mathfrak{F}}{\int_\mu{(|f|-n)^+\diff\omega}}=0.$$
The goal is to show that remark 2 holds. In order to do so I need to show that if some family satisfies definition (1) then it also satisfies definition (3).
My idea is to plug in the definition of limit and see how we need to guess. So if $\epsilon>0$ then I need to find a $N\in\mathbb{N}$ such that for all $n\geq N$ I have that $$\sup_{f\in\mathfrak{F}}{\int_\mu{(|f|-n)^+\diff\omega}}<\epsilon.$$ Definition (2) yields an $h'\in\integrableP$ such that $$\sup_{f\in\mathfrak{F}}{\int_\mu{(|f|-h')^+\diff\omega}}<\frac{\epsilon}{2}$$ holds. By triangle inequality I have $(|f|-n)^+\leq(|f|-h')^++(h'-n)^+$. The question thus boils down to asking if $\int_\mu{(h'-n)^+\diff\omega}<\frac{\epsilon}{2}$ holds for large $n\in\mathbb{N}$.
This is where I am stuck. I tried different approaches but nothing worked. This is what may help:
(a)
$$h'\in\integrable\implies\mu(\{h=\infty\})=0$$
(b)
$$\mu(\{h=\infty\})=0\implies\lim_{n\to\infty}\mu(\{h'\geq n\})=0$$
(c)
$$\mu(\Omega)<\infty$$
Note that $$(h'(x)-n)^+ = 0$$ for all $x$ with $h'(x) \leq n$. Thus
$$\int (h'-n)^+ \, d\mu = \int_{\{h' \geq n\}} (h'-n)^+ \, d\mu \leq \int_{\{h' \geq n\}} h' \, d\mu.$$
Applying the dominated convergence theorem we get
$$\lim_{n \to \infty} \int (h'-n)^+ \, d\mu = \int_{\{h'=\infty\}} h' \, d\mu=0$$
as $\mu(\{h'=\infty\})=0$.