Let $\mu$ be a probability measure on $X$, so that $\int_X \mu(dx) = 1$.

I have a family $\{f_i\}_{i=1}^{\infty}$ of functions $f_i: X \rightarrow \mathbb{R}_{\geq 0}$ such that $$ \displaystyle \sup_{i} \int_X f_i(x) \mu(dx) < \infty$$

Under which additional conditions I can say that they are Uniformly Integrable?

The sequence I have is general, of the kind $f_i(x) = g(z_i,x)$ where $g: \mathbb{R}^n \times X \rightarrow \mathbb{R}_{\geq 0}$ is locally bounded (and integrable for any $z \in \mathbb{R}^n$). Moreover, $z_i \in \mathbb{B}(z,\delta)$, i.e. the ball centered in $z \in \mathbb{R}^n$ with radius $\delta$. I can make such radius $\delta$ as small as desired.

How to conclude that such family is Uniformly Integrable (if it's the case)?

  • $\begingroup$ The subject is somewhat tricky, at least for me, so maybe this can help, it is a sort of an attempt to demonstrate what the uniform integrability is: websfog.blogspot.co.il/2012/09/probability-1.html $\endgroup$ – user47756 Nov 1 '12 at 5:30

http://en.wikipedia.org/wiki/Uniform_integrability. Of course, there are some obvious conditions, like monotonicity of the sequence $\{f_i\}$. Here by monotonicity it is meant that for any measurable $A\subset X$, $\|f_n|_A\|_{L^1}\leq \|f_m|_A\|_{L^1}$ whenever $n\geq m$. Check that link for sufficient conditions.

The most general sufficient conditions are hard to come up with, simply because there are easy counterexamples even in the nicest cases (i.e. where you're dealing with a sequence of, say, continuous functions). As far as particular cases: there would be too many to list. Maybe you can tell us more about the sequence you have in mind?

EDIT: The answer above, and most of the discussion below, pertains to the general case before the OP posted the extra conditions (i.e. the stuff about function $g$), and clarified what is meant by local boundedness of $g$. Below is the answer pertaining to the specific family $f_i(\cdot) = g(z_i, \cdot)$.

Case i.

If $g$ is locally bounded independently of the second component, that is, if for every $z\in \mathbb{R}^n$ there exists $\delta > 0$ such that for every $w\in\mathbb{B}(z, \delta)$ and almost every $x\in X$, $|g(w,x)|\leq M$, then the answer to the original question is obviously yes. Indeed, then there exists $\delta_0 > 0$ and $M > 0$ such that for all $w\in\mathbb{B}(z, \delta_0)$ (with $z$ as in the original question), and almost every $x\in X$, $|g(w, x)|\leq M$. In particular, if $\{z_i\}$ is a sequence in $\mathbb{B}(z,\delta_0)$, then $|f_i|\leq M$ almost surely on $X$ (as the OP said, $\delta$ in the original post can be taken arbitrarily small, in particular we may take $\delta = \delta_0$). But then obviously we have uniform integrability.

Case ii.

Now suppose we're dealing with a less trivial example. Assume that $X$ is a topological space. Suppose that $g$ is such that for any $(z,x)\in\mathbb{R}^n\times X$, there exists an open neighborhood $\mathbb{B}$ around $(z, x)$ in $\mathbb{R}^n\times X$ and $M > 0$, such that on $\mathbb{B}$, $|g|\leq M$ (that is, local boundedness of $g$ depends on both components). Then the following counterexample illustrates that uniform integrability generally doesn't hold.

Define $g$ as follows. Let $X = \mathbb{R}_{> 0}$. Let $g(y,x) = f_x(y)$, where $f_x(y) = 0$ on $(0, 1/\|x\|]\cup [1/\|x\| + \|x\|^{1/2}, \infty)$, and $f_x(y) = 1/\|x\|$ on $(1/\|x\|, 1/\|x\| + \|x\|^{1/2})$.

Now take $\delta < 1$, and consider $g$ on $\mathbb{B}(0, \delta)$. Pick any $v\in\mathbb{R}^n$ with $\|v\| = 1$ and define the sequence $\{f_i\}_{i\in \mathbb{N}}$ by $f_i(x) = g(v/i, x)$.

  • $\begingroup$ Can you please be more precise/rigorous about the "monotonicity"? Thanks! Of course, I've already read Wikipedia. $\endgroup$ – user693 Mar 29 '12 at 23:26
  • $\begingroup$ Sure, I just edited the answer to include the definition. By $f|_A$ I mean the restriction of $f$ to the subset $A$. $\endgroup$ – user2093 Mar 29 '12 at 23:31
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    $\begingroup$ Take $X = \mathbb{R}_{\geq 0}$ and consider the sequence $\{f_i\}$ given as follows. For $k\in\mathbb{N}$, define $f_k(x) = 0$ on $[0, k]\cup [k + 1/k, \infty)$, and $f_k(x) = k$ on $(k, k + 1/k)$. Why isn't this a counterexample (with respect to the Lebesgue measure, of course)? $\endgroup$ – user2093 Mar 30 '12 at 0:03
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    $\begingroup$ Absolute continuity would do the trick. The counterexample above is also a sequence of functions uniformly bounded in $L^1$ (the $L^1$-norm of each function is equal to one). $\endgroup$ – user2093 Mar 30 '12 at 0:52
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    $\begingroup$ Yes. The trick is to note that the functions are all bounded a.e. uniformly by the same constant. $\endgroup$ – user2093 Mar 30 '12 at 18:35

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