Let $(\Omega,\Sigma,\mu)$ be a finite measure space and let $F=\{f_{\alpha}:\Omega\to \mathbb{R}:\alpha\in I\}$ (where $I$ is some index set) be an arbitrary family of measurable real function on $\Omega$.
Let $\lim_{\lambda\to\infty}\int_{\{|f_{\alpha}|>\lambda\}}|f_{\alpha}|d\mu=0$,uniformly in $\alpha\in I$ then there exists a convex function $\Phi:\mathbb{R}\to\mathbb{R^+}$ such that $\Phi(0)=0, \Phi(-x)=\Phi(x)$ and $\frac{\Phi(x)}{x}\uparrow \infty$ as $x\uparrow \infty$, in terms of which $$C_{1}=\sup_{\alpha}\int_{\Omega}\Phi(f_{\alpha})d\mu<\infty .\tag{A}$$
$\textbf{Proof:}$ Let we have $\lim_{\lambda\to\infty}\int_{\{|f_{\alpha}|>\lambda\}}|f_{\alpha}|d\mu=0$,u niformly in $\alpha\in I$ is true. Choose $0<\lambda_{n}<\lambda_{n+1}\to \infty$ such that
$$\sup_{\alpha}\int_{\{|f_{\alpha}|>\lambda_{n}\}}|f_{\alpha}|d\mu<\beta_{n}; \hspace{0.5mm} n\geq 1 \tag{1}$$
where $\{\beta_{n},n\geq1\}$ is any sequence satisfying $\sum_{n}\beta_{n}<\infty$. Let $a_{0}=0$ and for $n\geq n+1$, let $a_n$ be the number of $\lambda_{n}$'s of $(1)$ which lie in the interval $[n,n+1).$ Set $a_{n}=0$ if there is no $\lambda_{n}$'s. Let $\phi(n)=\sum_{k=0}^{n}a_{k}$. Then $\phi(n)\uparrow \infty$ as $n\to \infty$ and let $\phi(t)=\phi(n),n\leq t<n+1,$ and set
$$\Phi(x)=\int_{0}^{|x|}\phi(t)dt\hspace{1cm} ;x\in\mathbb{R}$$.
Since $\phi(.)$ is increasing ,it follows that
$\Phi(x)=\Phi(-x)$,$\hspace{0.5mm}$ $\frac{\Phi(x)}{x}\geq\phi(k)\frac{x-k}{x}\uparrow \infty$ as $k<x$ and $k\to\infty.$ Moreover $\Phi(.)$ is a convex function that is $\Phi(\alpha x+\beta y)\leq \alpha\Phi(x)+\beta\Phi(y),0\leq \alpha,\beta\leq 1,\alpha+\beta=1.$ This follows from the fact that $\phi(.)$ is increasing and use the change of variable. To see that this $\Phi$ satisfies the condition $(A)$,
$$\int_{\Omega}\Phi(f_\alpha)d\mu=\sum_{n=1}^{\infty}\int_{\{n-1\leq|f_{\alpha}|<n\}}\Phi(|f_{\alpha}|)d\mu\leq \sum_{n=1}^{\infty}\Phi(n)\mu(n-1\leq|f_{\alpha}|<n)$$
$$\leq \sum_{n=0}^{\infty}[\Phi(n+1)-\Phi(n)]\mu(|f_{\alpha}|\geq n)\leq \sum_{n=0}^{\infty}\phi(n)\mu(|f_{\alpha}|\geq n).$$
Also,
$$\int_{\{|f_{\alpha}|\geq \lambda_{n}\}}|f_{\alpha}|d\mu=\sum_{r=\lambda_{n}}^{\infty}\int_{\{r\leq |f_{\alpha}|<r+1\}}|f_{\alpha}|d\mu\geq \sum_{r=\lambda_{n}}^{\infty}r\mu[r\leq|f_{\alpha}|<r+1]$$
$$\geq \sum_{r=\lambda_{n}}^{\infty}\mu[|f_{\alpha}|>r] \tag{B}$$
it follows from this on adding over $n$ that
$$\sum_{n=1}^{\infty}\phi(n)\mu[|f_{\alpha}|>n]=\sum_{n=1}^{\infty}\sum_{r=\lambda_{n}}^{\infty}\mu[|f_{\alpha}|>r]\leq \sum_{n=1}^{\infty}\int_{\{|f_{\alpha}|>\lambda_{n}\}}|f_{\alpha}|d\mu<\sum_{n=1}^{\infty}\beta_{n}<\infty.\tag{C}$$
Mu questions are:
(1) How we got $\frac{\Phi(x)}{x}\geq \phi(k)\frac{x-k}{x}$ as $k<x$ and $k\to \infty$?
(2) How it says that $\Phi(.)$ is convex , if I apply the definition of convexity to $\Phi(x)=\int_{0}^{x}\phi(t)dt$ then we have $\Phi(\alpha x+\beta y)=\int_{0}^{|\alpha x+\beta y|}\phi(t)dt$ then how we get from here the quantity $\alpha\Phi(x)+\beta \Phi(y)?$
(3) And I don't understand how we are getting $\textbf{(B)}$ and $\textbf{(C)}$.
Please, if someone explain this that will be great help for me. I am stuck here from 2-3 days. Thanks.