Let $\mu$ be a finite measure and let $(f_i)_{i\in I}$ be a sequence of functions such that $$ f_i\le (k g)^{1/2} \quad \forall i\in I, $$ where $k>0$ is a constant and $g\in L^1(\mu)$. I am trying to prove that $(f_i)_{i\in I}$ is a uniformly integrable over $E$, i.e. $$ \forall\varepsilon>0 \, \exists\delta>0 \text{ s.t. if } A\subseteq E \, \text{ with } \, \mu(A)\le \delta\implies \int_A |f_i| d\mu \le\varepsilon \quad\forall i\in I. $$
According to me, this is true simply because for any $i\in I$, $(f_i)$ is bounded by a function in $L^1$ which is uniformly integrable (by the absolute continuity of the Lebesgue integral).
Anyway, I am looking for a detailed $\varepsilon$-$\delta$ proof of this particular case, which I can not prove. Anyone could please help with that?
Without loss of generality, $k = 1$. Given $\varepsilon>0$, choose $\delta >0$ such that $$\int_A |g|\leq \varepsilon^2,\quad \mu(A)\leq \delta$$ Then $$\int_A |f_i|\leq\left\{ \int_Af_i^2\right\}^{1/2} \leq \left\{\int_A|g|\right\}^{1/2}\leq \varepsilon$$