3
$\begingroup$

Let $\mu$ be a probability measure on $\mathbb{R}^m$ (so $\int_{\mathbb{R}^m} \mu(d x) = 1$).

Let $f_i:\mathbb{R}^m \rightarrow \mathbb{R}_{\geq 0}$ be integrable functions and let also $\limsup_{i \rightarrow \infty} f_i$ be an integrable function.

Assume that

$$ \limsup_{i \rightarrow \infty} \int_{K} f_i(x) \ \mu(d x) \leq \ c \quad \text{ for all compact sets } K \subset \mathbb{R}^m.$$

Find under which additional assumptions (for instance on the $f_i$s and/or on $\int f_i$s) we also have:

$$ \limsup_{i \rightarrow \infty} \int_{\mathbb{R}^m} f_i(x) \ \mu(d x) \ \leq \ c$$

Notes: The measure in the two integrals is the same. What changes is the domain: from $K$ (compact, arbitrarily large) to $\mathbb{R}^m$. $c \in \mathbb{R}_{\geq 0}$.

  • 0
    I suspect it's true.2012-03-02
  • 1
    @Adam I don't know the site etiquette, but it strikes me as against it to ask a question, accept an answer, then change the question based on the answer (and unaccept answer), but I'll remove my answer as it no longer seems relevant.2012-03-02
  • 0
    I'm sorry about that.. I thought opening a new (closely-related) question would have been even worse.. By the way, I did appreciate your answers.2012-03-02
  • 0
    I'll undelete because of comments and note that it is no longer an answer.2012-03-02

2 Answers 2

8

This is not true in general.

Consider $\mathbb{R}$ with Lebesgue measure. Let $f_i(x)=i$ on the interval $[i,i+1]$. For any compact set $K$, for large $i$, $f_i$ will be $0$ on $K$, so your first limit is $0$. However, $\int_{\mathbb{R}}f_i(x)dx=i$, so your second limit is $\infty$.

EDIT

This was an answer to the original question, but is not relevant to current question. Perhaps the comments are useful though.

  • 0
    If $\limsup_{i\rightarrow \infty} f_i$ is integrable, does the result hold true?2012-03-02
  • 1
    Even if you made $f_i(x)=1$ on $[i,i+1]$ so that $\lim_{i \rightarrow \infty} f_i =0$, the result would still not hold (with say $c=1/2$).2012-03-02
  • 0
    i guess $\lim_{i \rightarrow \infty} f_i=0$ in the first example, but the integrals are blowing up.2012-03-02
  • 0
    Does the result hold if $\mu$ is a probability measure? Say, for instance, we have a Gaussian distribution. The first integral is still $0$. Is the second one, after taking the $\limsup_{i \rightarrow \infty}$ also $0$?2012-03-02
  • 1
    I think you could still have mass that escapes. I think this probably is true if your system is uniformly integrable (http://en.wikipedia.org/wiki/Uniform_integrability)2012-03-02
  • 0
    I modified the question according to your comments.. thanks2012-03-02
  • 0
    I'm trying to find a counterexample, but if the distribution is integrable then "the tail is quite small".. therefore I haven't found an example [I tried distributions like $e^{-x}$, $1/x^2$] in which there is some "mass escape"..2012-03-02
  • 0
    Let's say I have the probability density function: $p(x) = \frac{1}{x^2}$ if $x \geq 1$, $0$ otherwise, so that $\int_{\mathbb{R}} p(x) dx = 1$. Consider $f_i(x) = x^2 $ if $x \in [i,i+1]$, $0$ otherwise. Then: $$\displaystyle \limsup_{i \rightarrow \infty} \int_{K} f_i(x) p(x) dx = 0 $$ for any (arbitrarily large) compact set $K$, while $$ \limsup_{i \rightarrow \infty} \int_{\mathbb{R}} f_i(x) p(x) dx = 1.$$2012-03-03
  • 1
    Are you able to prove it's true if your random variables are uniformly integrable?2012-03-03
  • 0
    Hi ShawnD, thanks for the question. I'm not familiar with "uniformly integrability": can you make an example, please? Then, do you think my counterexample could be correct, at least according to the assumptions made so far? Thanks2012-03-04
  • 0
    I guess you mean: $$ \displaystyle \lim_{c \rightarrow \infty} \sup_{i} \int_{K_i(c)} f_i(x) p(x) dx = 0 $$ where $$ K_i(c) = \{ x: \ f_i(x) > c \}. $$ Anyway: NO I wasn't able to prove it. Suggestions?2012-03-06
0

After writing this answer I realized that i mistook $\limsup$ for $\sup$ - so this is not actually an answer but still might help finding one..


From $$ c \geq \sup_i \int_K f_i(x) d\mu(x) $$ for all compact subsets $K$ of $\mathbb{R}^m$ we follow that \begin{align} c \geq & \sup_{K\subset\mathbb{R}^m}\sup_i \int_K f_i(x) d\mu(x) \\ = & \sup_i \sup_{K\subset\mathbb{R}^m} \int_K f_i(x) d\mu(x)\\ = & \sup_i \int_{\mathbb{R}^m} f_i(x) d\mu(x) \end{align}