I'm trying to prove below theorem mentioned in this comment. This theorem allows us to generalize this property from finite measure spaces to $\sigma$-finite measure spaces.
Theorem: Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space. There is a finite measure $\nu$ on $(X, \mathcal A)$ such that $\mu$ is absolutely continuous w.r.t. $\nu$.
Could you have a check on my bellow attempt?
Proof: Let $(X_m)$ be a countable measurable partition of $X$ such that $\mu(X_m) <\infty$. We define a measure $\mu_n$ by $$ \mu_n (A) := \mu(A \cap X_m) \quad \forall A \in \mathcal A. $$
Then $\mu_n$ is finite and supported on $X_n$. We define a measure $\nu_n$ by $$ \nu_n(A) := \frac{\mu_n(A)}{2^{n}\mu_n(X_n)}. $$
Then $\mu_n \ll \nu_n$ and $\nu_n(X) = 2^{-n}$. By Radon–Nikodym theorem, there is a measurable function $f_n:X \to [0, \infty)$ such that $$ \mu_n(A) = \int_A f \mathrm d \nu_n(A) \quad \forall A \in \mathcal A. $$
Because $\nu_n$ is supported on $X_n$, we can assume $f_n=0$ on $X \setminus X_n$. Let $f := \sum_n f_n$. Then $f$ is the Radon–Nikodym derivative of $\mu$ w.r.t. $\nu :=\sum_n \nu_n$. Notice that $$ \nu(X) = \sum_n \nu_n(X_n) = \sum_n 2^{-n} = 1. $$
This completes the proof.