Approximation of $\sigma$-finite Borel measures by equivalent finite measures

Question

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 $\mathbb{Q}\cap(0,\infty)$.

Answer 1

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 $1$.

Answer 2

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 $\mu_n \uparrow \mu$.