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Let $\{\mu_k\}_{k>1}$ be a sequence of radon measures on $\mathbb{R}^n$ satisfying $\sup_k\mu_k(K)<\infty$ for each compact set $K\subset \mathbb{R}^n$. Prove that if $\sup_k\mu_k(\mathbb{R}^n)=∞$ then there still exists a sub sequence $\{\mu_{k_j}\}$ and a radon measure $\mu$ such that $\mu_{k_j}$ converge weakly to $\mu$.

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  • $\begingroup$ What are your thoughts so far? $\endgroup$ – Christopher A. Wong Apr 1 '14 at 21:55
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    $\begingroup$ I can prove this sequence has weakly convergent subsequence when $sup\mu_k(\mathbb{R}^n)<\infty$. Now we can define a sequence of sequence of radon measure by restriction to ball with radius $r$, i.e. $\mu_k|_{B(0,r)}=\mu^r_k$. Therefore, for each r, $\mu^r_k$ has weakly convergent subsequent $\mu^r_{k_s}$ to Radon measure $\mu^r$. Now I want to use diagonal argument and I need help!! The problem is there is no relation between this collection of subsequnce!!! $\endgroup$ – John Apr 1 '14 at 22:11
  • $\begingroup$ To continue with your approach, you could try using a diagonal argument by showing that it suffices to consider all balls of rational radius at rational centers. $\endgroup$ – Christopher A. Wong Apr 1 '14 at 22:16

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