Uniform integrability of a function in $L^1$ A collection of  functions $(\phi_i)_{i\in I}\in L^1(\mu)$ is called uniformly integrable if given $\epsilon>0$ there exists $\delta>0$ such that :
$$\int_E|\phi_i|d\mu<\epsilon~~~~\forall E:\mu(E)<\delta; \forall i\in I$$
Now the question is to prove that collection with exactly one element is uniformly integrable....
I mean given $f\in L^1$ and $\epsilon>0$ we need to produce $\delta>0$ such that 
$$\int_E|f|d\mu<\epsilon~~~~\forall E:\mu(E)<\delta;$$
What I have tried so far is as follows :
As $|f|$ is a positive measurable function there exist a sequence of simple functions converging to $f$ point wise...
Given $\epsilon>0$ there exists a simple function $s(x)$ such that 
$$\int_X |f|d\mu\leq \int_X s d\mu+\epsilon$$
I am not sure what should be the next step but if at all it is true I would like to write this as 
$$\int_E |f|d\mu\leq \int_E s d\mu+\epsilon ~~\text{ which holds} ~~ \forall E\subset X$$
If this is true then I have 
$$\int_E |f|d\mu\leq \int_E s d\mu+\epsilon$$
As $s$ is simple hence bounded and thus for some $M>0$ we have $s(x)<\leq M\forall x\in X$
i.e.,  $$\int_E |f|d\mu\leq \int_E s d\mu+\epsilon<M\mu(E)+\epsilon$$
Now I need to choose $\delta$ such that $\mu(E)<\delta$ imply $M\mu(E)+\epsilon<\epsilon $
this does not make sense so i replace all my $\epsilon$ in above calculation with $\dfrac{\epsilon}{2}$ except the last one.. i.e., 
I need to choose $\delta$ such that $\mu(E)<\delta$ imply $$M\mu(E)+\dfrac{\epsilon}{2}<\epsilon \Rightarrow M\mu(E)<\dfrac{\epsilon}{2}\Rightarrow \mu(E)<\dfrac{\epsilon}{2M}$$
Now I choose $\delta$ to be $\dfrac{\epsilon}{2M}$
I hope what I have done is partially true... I expect someone to check this and let me know if there are any mistakes..
EDIT : I assumed $$\int_X |f|d\mu\leq \int_X s d\mu+\epsilon \Rightarrow \int_E |f|d\mu\leq \int_E s d\mu+\epsilon ~~\text{ which holds} ~~ \forall E\subset X$$.. I am asking if this is true under some conditions.. This is not true in general... 
Please help me to  make this perfect..
 A: Let $s$ integrable and $\varepsilon$ such that $s\leqslant|f|$ on $X$ and $\displaystyle\int_X|f|\leqslant\varepsilon+\int_Xs$. Then, for every measurable $E\subseteq X$,  $|f|-s\geqslant0$ on $X\setminus E$ hence $\displaystyle\varepsilon\geqslant\int_X|f|-s=\int_E|f|-s+\int_{X\setminus E}|f|-s\geqslant\int_E|f|-s$ , which implies $\displaystyle\int_E|f|\leqslant\varepsilon+\int_Es$.
A: The proposition is trivial if the function $f$ is bounded.  So assume that $f_n(x) = n$ if $f(x) \leq n$ and $f_n(x) = 0$ otherwise.  Then each $f_n$ is bounded and $f_n \to f$ pointwise so by the Monotone convergence theorem $\int_E f_n \to \int_E f$.  So given $\epsilon > 0$ there exists an $N$ such that $\int_E f - \int_E f_N < \epsilon/2$.  Choose $\delta < \epsilon/2N$.  If $m(A) < \delta$, we have that $\int_A f = \int_A f - f_N + f_N < \int_E (f - f_N ) + Nm(A) < \epsilon$ as needed.
A: Maybe a different approach would be this:
Since $f \in L^1(\mu)$ we know by standard measure theory that $|f|<\infty$ [$\mu$] a.e. Consider $A_n \equiv \{|f|>n\}$ and set $f_n \equiv |f|\chi_{A_n}$, then clearly $f_n \leq |f|$ and since $\{|f|=\infty\}$ has measure 0, we have $f_n \to 0$ as $n \to \infty$. Thus by the Dominated Convergence Theorem $\int f_n d\mu = \int |f| \chi_{A_n} d\mu \to 0$ as $n \to \infty$. Thus for $\epsilon>0$ there exists $N_\epsilon$ big enough such that $\int |f| \chi_{A_{N_{\epsilon}}} d\mu<\epsilon$, implying uniform integrability.
