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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 $\mathbb{R}^m$. $c \in \mathbb{R}_{\geq 0}$.

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I suspect it's true. – Michael Hardy Mar 2 '12 at 4:23
@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. – ShawnD Mar 2 '12 at 20:00
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. – Adam Mar 2 '12 at 20:14
I'll undelete because of comments and note that it is no longer an answer. – ShawnD Mar 2 '12 at 20:15

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 $\infty$.


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

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If $\limsup_{i\rightarrow \infty} f_i$ is integrable, does the result hold true? – Adam Mar 2 '12 at 5:39
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 $c=1/2$). – ShawnD Mar 2 '12 at 5:44
i guess $\lim_{i \rightarrow \infty} f_i=0$ in the first example, but the integrals are blowing up. – ShawnD Mar 2 '12 at 5:58
I think you could still have mass that escapes. I think this probably is true if your system is uniformly integrable ( – ShawnD Mar 2 '12 at 6:25
Are you able to prove it's true if your random variables are uniformly integrable? – ShawnD Mar 3 '12 at 23:13

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