Let $f$ be differentiable s.t. $f'>0$ everywhere and $\nu = f_{\sharp} \mu$. Is $\nu \ll \mu$?

Question

Let $\mu$ be a finite Borel measure on $\mathbb R$ such that $\mu$ is absolutely continuous (a.c.) w.r.t. Lebesgue measure $\lambda$. Let $f:\mathbb R \to \mathbb R$ be differentiable such that $f'(x)>0$ for all $x \in \mathbb R$. Let $\nu$ be the push-forward of $\mu$ by $f$, i.e., $\nu = f_{\sharp} \mu$. I proved that $\nu$ is also a.c. w.r.t. $\lambda$. Now we assume the p.d.f. of $\mu, \nu$ are $p_\mu, p_\nu$ respectively. Then $$ p_\nu = \frac{p_\mu \circ f^{-1} }{f' \circ f^{-1}}. $$

We are now interested in below result, i.e.,

Theorem: $\nu$ is a.c. w.r.t. $\mu$

Proof: We now assume $\nu$ is a.c. w.r.t. $\mu$ and $p := \frac{\mathrm d \nu}{\mathrm d \mu}$. Fix a Borel set $A$ of $\mathbb R$. Then $$ \int_A p \mathrm d \mu = \nu(A) = \int_A p_\nu \mathrm d \lambda. $$

Then $$ \int_A p p_\mu \mathrm d \lambda = \int_A \frac{p_\mu \circ f^{-1} }{f' \circ f^{-1}} \mathrm d \lambda. $$

The proof is completed by picking $$ p := \frac{p_\mu \circ f^{-1} }{p_\mu (f' \circ f^{-1})}. $$

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1. Could you confirm if my proof is fine or contains logical mistakes?
2. Are there other proofs that are less "mechanical" then mine?
3. Is the absolute continuity of $\mu$ w.r.t. $\lambda$ essential for that of $\nu$ w.r.t. $\mu$?

Answer 1

As for your proof, it seems fine (note that if $p_\mu$ is $0$ we can define $p$ to be $0$ (or anything) because the set $\{x\in \mathbb{R} | p_\mu(x)=0\}$ has $\mu$-measure $0$), at least assuming the previous proofs to be correct.

As for your second question, sadly I have none yet.

As for the third question, it seems to be the case. Consider $x_0\in \mathbb{R}$ and $\delta_{x_0}$ the Dirac measure given by $\delta_{x_0}(A) = \chi_A(x_0)$. Then if we choose $\mu = \delta_{x_0}$, $f(x)=x+1$ we get that $\nu=\delta_{x_0-1}$ which is not absolutely continuous with respect to $\mu$ (for example, $\mu(\{x_0+1\})=0$ but $\nu(\{x_0+1\})=1$).