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..
Best Answer
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 $\displaystyle\int_E|f|\leqslant\varepsilon+\int_Es$.