Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space. Previously, I proved that
Lemma There is a finite measure $\nu$ on $(X, \mathcal A)$ such that $\mu$ has a density $f:X \to [0, \infty)$ w.r.t. $\nu$.
Now I would like to strengthen above result, i.e.,
Theorem: There is a finite measure $\nu$ on $(X, \mathcal A)$ such that $\mu$ has a positive density $f:X \to (0, \infty)$ w.r.t. $\nu$.
Could you have a check on my below attempt?
Proof: Let $(f, \nu)$ be given by the Lemma. Let $N := f^{-1} (0)$. Because $f$ is measurable, $N \in \mathcal A$. This means $N$ is a $\mu$-null set. Let $N^c := X \setminus N$ and we define a measure $\lambda$ by $$ \lambda(A) := \nu (A \cap N^c) \quad \forall A \in \mathcal A. $$
Then $\lambda$ is concentrated on $N^c$. We define $g:X \to (0, \infty)$ by $$ g(x) := \begin{cases} f(x) &\text{if} &x \in N^c \\ 1 &\text{if} &x \in N. \end{cases} $$
Then
- $\lambda$ is a finite and $g$ measurable.
- $f$ and $g$ agree on $N^c$.
- $\nu$ and $\lambda$ agree on the sub $\sigma$-algebra of $N^c$.
- $\lambda = 0$ on the sub $\sigma$-algebra of $N$.
As such, for every $A \in \mathcal A$ we have $$ \mu (A) = \int_A f \mathrm d \nu = \int_{A \cap N^c} f \mathrm d \nu = \int_{A \cap N^c} g \mathrm d \lambda = \int_A g \mathrm d \lambda. $$
As such, the pair $(g, \lambda)$ satisfies our requirement. This completes the proof.
