# The absolute continuity of push-forward measure

Let $$X$$ be a real-valued random variable on $$\mathbb R$$, and $$f:\mathbb R \to \mathbb R$$ differentiable such that $$f'(x)>0$$ for all $$x \in \mathbb R$$. Let $$Y := f(X)$$. Let $$\mu_X, \mu_Y$$ be the distributions of $$X, Y$$ respectively. Then $$\mu_Y = f_{\sharp} \mu_X$$. Let $$F_X, F_Y$$ be the c.d.f. of $$X, Y$$ respectively. At page $$14$$ of this lecture note, the author said that

Theorem: If $$\mu_X$$ is absolutely continuous w.r.t. Lesbesgue measure $$\lambda$$, then so is $$\mu_Y$$

My attempt: Clearly, we have $$F_Y (t) = F_X \circ f^{-1} (t)$$. Let $$A$$ be a Borel set such that $$\lambda(A) = 0$$. Because $$\mu_X \ll \lambda$$, we get $$\mu_X (A) =0$$. We have $$\mu_Y (A) = \mu_X(f^{-1} (A))$$. It suffices to prove $$f^{-1} (A)$$ is a $$\lambda$$-null set.

Could you shed some light on how to finish the proof?

Update: I have found a related result here. However, it requires $$f$$ to be continuously differentiable, i.e., if $$f\in C^1$$ and $$\{f' = 0\}$$ is $$\lambda$$-null then $$f^{-1} (A)$$ is also $$\lambda$$-null.

• I feel like Baire's category theorem could be somehow useful here... Sep 27, 2022 at 10:53
• @Masacroso In the link you gave, the function is differentiable everywhere, while in our case $F_X$ is differentiable almost everywhere. Please see here for my failed attempt to prove that $F_Y = F_X \circ f^{-1}$ is absolutely continuous. Sep 27, 2022 at 21:04
• @Masacroso I could not see how Lemma 7.20 in that book helps us. Could you elaborate more? Maybe Theorem 7.21 is relevant, but the function $F$ in that theorem is assumed to be absolutely continuous. Sep 27, 2022 at 21:05
• @Masacroso The composition $a \circ b$ of 2 absolutely continuous (a.c.) functions $a,b$ is not necessarily a.c. However, if in addition $b$ is (not necessarily strictly) monotone, then $a \circ b$ is a.c. Could you confirm if my understanding is correct? Sep 27, 2022 at 21:36
• @Akira I dont know. My reasoning was not correct before, sorry I didn't paid enough attention to this question. I would try to come with a definitive answer (I deleted my previous comments as they seems wrong) Sep 27, 2022 at 21:42

I formulate @Masacroso's idea as follows. First, we need 3 lemmas.

• Lemma 1: If $$g:\mathbb R \to \mathbb R$$ is monotone and differentiable, then $$g$$ is absolutely continuous (a.c.).

• Lemma 2: Let $$F:\mathbb R \to \mathbb R$$ and $$g:\mathbb R \to \mathbb R$$ be both a.c. where $$g$$ is (not necessarily strictly) monotone. Then $$F \circ f : \mathbb R \to \mathbb R$$ is a.c.

• Lemma 3: A finite measure $$\mu$$ on Borel subsets of $$\mathbb R$$ is a.c. w.r.t. Lebesgue measure if and only if the associated function $$F(x) := \mu((-\infty, x])$$ is a.c.

Clearly, $$f^{-1}$$ is monotone, and differentiable by inverse function theorem, then $$f^{-1}$$ is a.c. by Lemma 1. By Lemma 3, $$F_X$$ is a.c. By Lemma 2, $$F_Y = F_X \circ f^{-1}$$ is a.c. Then $$\mu_Y$$ is a.c. by Lemma 3.

Update: I use the following version of inverse function theorem (IFT) (at page 306 of Amann's Analysis I), i.e.,

Let $$X$$ be an open subset of $$\mathbb{K} \in \{\mathbb{C}, \mathbb{R}\}, f: X \rightarrow \mathbb{K}$$, and $$Y:=f(X)$$. Let $$f$$ be injective and consider the inverse $$f^{-1}: Y \rightarrow X$$ of $$f$$. Suppose that $$f$$ is differentiable at $$a \in X$$, and $$f^{-1}$$ is continuous at $$b:=f(a) \in Y$$. Then $$f^{-1}$$ is differentiable at $$b$$ if and only if $$f^{\prime}(a) \neq 0$$. In this case, $$\left(f^{-1}\right)^{\prime}(b)=1 / f^{\prime}(a)$$.

To use IFT we need to prove that $$f^{-1}$$ is continuous. However, this follows from invariance of domain theorem.