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Let $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d),\mu)$ be a $\sigma$-finite Borel measure on $d$-dimensional Euclidean space. Can one always construct a sequence of finite equivalent measures $\left\{\mu_n\right\}_{n \in \mathbb{N}}$ on $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$ such that $$ \lim_{n \rightarrow \infty} \frac{d\mu_n}{d\mu} \uparrow 1. $$

Motivation: This is true for both the Lebesgue measure and the counting measure on $\mathbb{R}^d$ by taking $\frac{d\mu_n}{d\mu}(x)\triangleq \begin{cases} e^{-q_n\min\{\|x\|,\frac1{\|x\|}\}}&:\, x\neq 0\\ 1 &: \, x =0 \end{cases} $, where $q_n$ runs over $\mathbb{Q}\cap(0,\infty)$.

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Since $\mu$ is $\sigma$-finite there is a $\mu$-integrable function $f$ with $0<f\le 1$. For an increasing sequence $A_n$ with $\bigcup_{n\in\mathbb N} A_n=\mathbb R^d$ define densities $f_n=I_{A_n}+fI_{A_n^c}$. Then $\mu_n=f_n\cdot \mu$ are equivalent finite measures whose densities converge to $1$.

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  • $\begingroup$ But how to show that such a function exists (ie the bounded away from 0 part)? $\endgroup$ – AIM_BLB Jan 10 at 10:07
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    $\begingroup$ If $A_n$ are as in the answer set $B_1=A_1$, $B_n=A_n\setminus A_{n-1}$, and $f=\sum\limits_{n=1}^\infty c_n 1_{B_n}$ with suitable $c_n>0$, so that $\int f d \mu= \sum\limits_{n=1}^\infty c_n \mu(B_n)<\infty$. $\endgroup$ – Jochen Wengenroth Jan 10 at 10:17
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    $\begingroup$ Note that this has nothing to do with the topology of $\mathbb R^d$. It holds for every $\sigma$-finite measure space. $\endgroup$ – Jochen Wengenroth Jan 10 at 10:18
  • $\begingroup$ Yes I had noticed that; thanks :) $\endgroup$ – AIM_BLB Jan 10 at 10:20
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Let $(E,\cal{E},\mu)$ be a measure space and $\mu$ a not finite $\sigma$-finite measure on $\cal{E}$. Then there is a sequence $E_n$ of $E_n \in \cal{E}$ with $E_n \uparrow E$ and $0 < \mu(E_n) < \mu(E_{n+1}) < \infty$. Let $a_1 := \mu(E_1)$ and $a_n := \mu(E_n \setminus E_{n-1})$ for $n > 1$. Let $\mu_n(A) := \mu(A \cap E_n) + \sum_{k=n+1}^\infty 2^{-k} / \mu(E_k \setminus E_{k-1}) \cdot \mu(A \cap (E_k \setminus E_{k-1}))$ for $A \in \cal{E}$. Then each $\mu_n$ is finite, equivalent to $\mu$ and $\mu_n \uparrow \mu$.

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  • $\begingroup$ Edited after reading Jochen Wengenroths answer. I ommited the topological assumtions. $\endgroup$ – Dieter Kadelka Jan 10 at 12:20

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