Uniform integrability of a function in $L^1$

Question

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..

Answer 1

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

Answer 2

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.

Answer 3

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.