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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'll undelete because of comments and note that it is no longer an answer.2012-03-02

2 Answers 2

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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.

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    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
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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}