Suppose that $g: \mathbb{R} \rightarrow \mathbb{R}$ is continuously differentiable and $\lambda$ is the Lebesgue measure.
What is the measure $\lambda_g$ where $\lambda_g(B) = \lambda(g^{-1}(B))$ is the pushforward measure for $g$.
Attempt at a Solution
This question seems like the answer will be that $\lambda_g(B) = \int_B |g'(x)| \lambda(dx)$. However, attempting to show this fact directly from the definition seems difficult. For example, if we could assume that $g$ was, say, monotone we could define the measure by noticing that
$\lambda_g((a,b)) = \lambda(g^{-1}(a,b)) = \lambda((g^{-1}(a), g^{-1}(b))) = g^{-1}(b) - g^{-1}(a)$ but this doesn't line up with my earlier idea.
Could it be then that $\lambda_g(B) = \int_B|g^{-1}{'}(x)| \lambda(dx)$ then?