Problem statement: For a measure space $ (\Omega, \mathcal{F}, \mathrm{μ})$ : ${μ} $ is positve and finite, show the following: The set function
$\nu(A) = \int_Af d\mu$ , $A \in \mathcal{F}$
defines a new positive measure on $\mathcal{F}$ equivalent to $\mathrm{μ}$, iff ${f}$ is strictly positive $\mathrm{μ}$-a.e and $\mathrm{μ}$-integrable.
My question: The forward direction obviously holds true from the Radon-Nikodym theorem. I'm having trouble proving the last tip of the reverse direction. Starting with a funcion ${f}$ as described above, It's not too hard to show that ${ν}$ is indeed a measure. It's also trivial that $\nu \ll \mu$. But how can we rigorously prove that $\mu \ll \nu$ ? I'm sure we've all used the argument that when for a strictly positive ${f}$, it holds:
$\int_Af d\mu = 0$ , then $\Rightarrow$ $\mu(A) = 0 $,
but what would be a formal and stringent proof of this property/claim ?
Let $f$ be strictly positive, $A_1 := f^{-1} [1, \infty)$ and $A_n := f^{-1}\left[\frac{1}{n}, \frac{1}{n - 1}\right)$ for $n \geq 2$.
Then we have $$ \mu(A) = \int_{A}{\mu(dx)} = \int_{A}\mu(dx) \frac{f(x)}{f(x)} = \int_{A}\mu(dx) \frac{f(x)}{f(x)} \sum_{n = 1}^{\infty}{1_{A_n}}(x) \\ = \sum_{n = 1}^{\infty} \int_{A}{\mu(dx) \frac{f(x)}{f(x)} 1_{A_n}}(x) \leq \sum_{n = 1}^{\infty} \int_{A}{\mu(dx) f(x) \frac{1}{\frac{1}{n} }} = \sum_{n = 1}^{\infty}{(n \cdot 0)} = 0. $$
An alternative proof by contradiction would be the following. Suppose that $\mu(A) > 0$. Then $$ \int_{A}{\mu(dx) f(x)} \geq \int_{A}{\mu(dx) \, f(x) \, 1_{f \geq \frac{1}{n}}(x)} \geq \frac{1}{n} \mu\left(A \, \cap \, \{f \geq \frac{1}{n}\} \right) > 0 $$ for $n \in \mathbb{N}$ big enough, because $$ \mu(A) = \lim_{n \to \infty} \mu\left(A \, \cap \, \{f \geq \frac{1}{n}\}.\right) $$