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

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

Since $$\mu$$ is $$\sigma$$-finite there is a $$\mu$$-integrable function $$f$$ with $$0. For an increasing sequence $$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$$.
• If $A_n$ are as in the answer set $B_1=A_1$, $B_n=A_n\setminus A_{n-1}$, and $f=\sum\limits_{n=1}^\infty c_n 1_{B_n}$ with suitable $c_n>0$, so that $\int f d \mu= \sum\limits_{n=1}^\infty c_n \mu(B_n)<\infty$. – Jochen Wengenroth Jan 10 at 10:17
• Note that this has nothing to do with the topology of $\mathbb R^d$. It holds for every $\sigma$-finite measure space. – Jochen Wengenroth Jan 10 at 10:18
Let $$(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$$.