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In a measure space with finite total measure, a family $A$ of r.v's is called U.I if

$$ \lim_{N\to\infty}\sup_{X\in A} \int_{|X|>N} |X|\,d\mu=0 $$

I have some doubts on equivalance of this definition in $L^1$ space. The following is equivalent to U.I in $L^1$:

  1. A is bounded subset of $L^1$.
  2. For every $\epsilon>0$ there exists $\delta>0$ such that for any measurable $E$ if $\mu(E)\le \delta$ then $\int_E |X| d\mu\le \epsilon$ for any $X\in A$.

The following proof is given:

Proof: Suppose $A$ is U.I. Then for any measurable $E$,

$$ \int_E |X| = \int_{E\cap \{|X|>N\}} |X| + \int_{E\cap \{|X|\le N\}} |X|:=B+C $$

Given $\epsilon>0$ we can choose $N$ such that $B<\epsilon /2$ and obviously $C\le N\mu (E)$. Thus we may choose $\delta = \frac{\epsilon}{2N(\mu(\Omega)+1)}$.


And I'm not sure why we can't simply choose $\delta=\epsilon/2N$? And why do we need the condition that $\Omega$ is a finite measure space? I don't particularly see any necessity for it, unless I'm missing something about last line. Probably only place where it might matter is stating $C\le N\mu(E)$, since it might be that $\mu(E)=\infty$. But we are controlling size of $E$ anyway, aren't we?

Edit: It seems for condition $1$ to hold, $E$ must be of finite measure so total measure must be finite, as we only control size of $E$ for proof of $2$.

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    $\begingroup$ Aren't you confusing uniform continuity with uniform integrability? $\endgroup$ Jun 4 '16 at 2:02
  • $\begingroup$ I don't quite understand what you mean, for which part do you think I'm talking about uniform continuity? It might be that I don't know what that is for a family of r.v, I only encountered that concept for a single function. $\endgroup$
    – user340297
    Jun 4 '16 at 2:06
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    $\begingroup$ You are right; there is nothing wrong about choosing $\delta = \epsilon/(2N)$. So, uniform integrability implies condition 2, but in order to show condition 1 (i.e. boundedness in $L^1$) you need the finiteness of the measure. $\endgroup$
    – saz
    Jun 4 '16 at 6:08
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It seems that the choice $\delta=\varepsilon/(2N)$ works well.

Finiteness of the measure space is used to show boundedness in $\mathbb L^1$ of a uniformly integrable family. Indeed, if the measure space is infinite, then a constant function is not integrable. Therefore, we cannot show that the family $\left(\int \left|X\right|\mathbf 1\{\left|X\right|\lt R\}\right)_{X\in A}$ is bounded. For example, assume that we work with the real line with Lebesgue measure and $X_n=\mathbf 1([n,2n))$. This family would be uniformly integrable if we consider the standard definition, but this is not bounded in $\mathbb L^1$.

What can be misleading in the statement in the opening post is that we talk about random variable, which supposes that we work with a probability space. In particular, there is no need to specify that the measure is finite.

Note that the concept of uniform integrability can be extended to infinite measure spaces. A family of functions $\left(f_i\right)_{i\in I}$ is uniformly integrable if for each positive $\varepsilon$, we can find an integrable function $g$ such that $$\sup_{i\in I}\int\left|f_i\right|\mathbf 1\left\{\left|f_i\right|\gt g\right\}\lt \varepsilon.$$

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