Setup
Given measurable spaces $(X_1,\Sigma_1)$ and $(X_2,\Sigma_2)$, a measurable mapping $f:X_1 \to X_2$ and a measure $\mu: \Sigma_1 \to [0,\infty]$, the push-forward of $\mu$ is defined to be the measure $f_{*}\mu:\Sigma_2 \to [0, \infty]$ given by
$$
(f_{*} \mu) (B) = \mu \left( f^{-1} (B) \right) \quad \mbox{ for } B \in \Sigma_{2}.
$$
Question
If $f$ is bijection, $f^{-1}$ is also measurable and $(X_1,\Sigma_1)=(X_2,\Sigma_2)$ are one and the same, then does this imply that
$$
\mu<<f_{*}\mu<<\mu.
$$
That is, does this imply that $\mu$ and its push-forward under $f$ are equivalent measures? Intuitively this seems to be true but I'm not certain if it is indeed the case.
Best Answer
The measures need not be equivalent. For example, consider any set $X$ with the Dirac measure $\delta_x$ centered at $x \in X$. Given any bijection $f:X \to X$, the pushforward measure $f_{\ast} \delta_x$ is simply $\delta_{f(x)}$, the Dirac measure centered at $f(x)$. Thus the measures are not equivalent unless $f(x)=x$, in which case they are equal.
An interesting question is: can we find "natural" conditions under which $f_{\ast} \mu$ and $\mu$ are equivalent? I don't know how to make this precise, but my impression is that this situation is the exception rather than the rule. Here are some thoughts about this, which are very far from an exhaustive picture.
Let $(X,\Sigma)$ be a measurable space, $\mu$ a measure on $X$ and $f:X \to X$ be a measurable map. $\mu$ and $f_{\ast}\mu$ are equivalent if and only if they have the same null sets. This means that $\mu(A) =0 \Leftrightarrow \mu( f^{-1}(A)) =0$. When $f$ has a measurable inverse, this can be restated as: $$ \mu(A) = 0 \qquad \Longrightarrow \qquad \mu(f(A))=0 \quad \text{and} \quad \mu(f^{-1}(A))=0 $$ This means that both $f$ and $f^{-1}$ satisfy Luzin's property N, that is they send sets of measure $0$ to sets of measure $0$.
Even in the case where $X= \mathbb{R}^n$ and $\mu$ is the Lebesgue measure, there is no "easy" characterization of maps with property N. Examples of such maps are locally Lipschitz maps (such as $C^1$ maps) and absolutely continuous maps in dimension $1$. (See here or here for details about this.)
However, one restrictive result is that maps of an interval which are continuous and have property N are differentiable on a set of positive measure (this is problem 5.8.57 of Bogachev's Measure Theory). We see that even in this simple situation the condition that $\mu$ and $f_{\ast} \mu$ be equivalent is rather restrictive, since "most" continuous functions are nowhere differentiable. (But then again, most measurable maps are discontinuous...)