some weak conditions about Vitali convergence theorem As we have known,the Vitali convergence theorem is stated:
Let $(X,\mathbb{M},\mu)$ be a positive measure space.If 
(i)$\mu(X)<\infty$;
(ii)$\{f_n\}$ is uniformly integrable;
(iii)$f_n(x)\to f(x)~~a.e.as~~n\to\infty$;
(iv)$|f(x)|<\infty~~a.e$;
then $f\in L^1(\mu)$ and
$$\lim_{n\to\infty}\int_X |f_n-f|d\mu=0.$$
Call a set $\Phi\subset L^1(\mu)$ uniformly integrable if to each $\varepsilon>0$ corresponds a $\delta>0$ such that $$|\int_E fd\mu|<\varepsilon$$
whenever $f\in\Phi$ and $\mu(E)<\delta$.
So I want to give some weaker conditions to verify whether the Vitali theorem is still valid or not.
First,if $\mu$ is Lebesgue measure on $(-\infty,\infty)$,and moreover,$\{\parallel f_n\parallel_1\}$ is assumed to be bounded.How about the Vitali theorem?
Second,can we omit the hypothesis (iv) when we consider the Lebesgue measure on a bounded interval?And can this be extended to finite measures?
Any help will be appreciated.

Here is the strongest version of Vitali's theorem (from O. Kavian, Introduction à la théorie des points critiques, Springer, 1993)
Definition. A sequence $\{f_n\}_n$ in $L^1(\Omega)$ is equi-integrable if the following condition is satisfied: for every $\varepsilon>0$ there exists a measurable set $A$ of finite measure and there exists $\delta>0$ such that


*

*for every $n \geq 1$, $\int_{\complement A} |f_n|<\varepsilon$;

*for every measurable $E \subset \Omega$ whose measure is less than $\delta$, there results $\int_E |f_n|<\varepsilon$ for every $n \geq 1$.


Theorem. Let $\{f_n\}_n$ be  a sequence in $L^1(\Omega)$ that converges almost everywhere to some measurable function $f$. The sequence $\{f_n\}_n$ converges to $f$ in $L^1(\Omega)$ if and only if it is equi-integrable.
