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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..

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  • $\begingroup$ To pass from a bound on the integral over $X$ to a bound on the integrals over every $E$ is direct if one assumes (and one can assume) that the simple functions $s$ are such that $s\leqslant|f|$. $\endgroup$ – Did Jul 28 '14 at 8:11
  • $\begingroup$ Quote: "As |f| is a positive measurable function there exist a sequence of simple functions converging to f point wise"... Indeed, and every lectures about integration theory will tell you that one can even choose these simple functions such that... $\endgroup$ – Did Jul 28 '14 at 8:46
  • $\begingroup$ Let me suggest to check in your textbook the theorem which guarantees that one can approach any integrable nonnegative function by a nondecreasing sequence of step functions. $\endgroup$ – Did Jul 28 '14 at 9:20
  • $\begingroup$ It is, under the condition that $s\leqslant|f|$ (as stated in my first comment). $\endgroup$ – Did Jul 28 '14 at 12:49
  • $\begingroup$ but that holds for a fixed $E$ right??? please.... i do not really know why i am not able to understand what you are saying... i would delete all those comments and let me start from beginning... $\endgroup$ – user87543 Jul 28 '14 at 12:56
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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 $\displaystyle\int_E|f|\leqslant\varepsilon+\int_Es$.

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  • $\begingroup$ Oh... I did not realize this.... Thank you :) $\endgroup$ – user87543 Jul 28 '14 at 13:15
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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.

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  • $\begingroup$ I think this is what i have done.... how is this different from what i have done... $\endgroup$ – user87543 Jul 28 '14 at 12:09
  • $\begingroup$ Your definition of $f_n(x)$ does not look right; perhaps you meant $f_n(x)$ to be $min\{ n, f \}$? $\endgroup$ – Wham Bang Shang-a-Lang Nov 4 '19 at 19:04
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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.

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