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.

  • $\begingroup$ I feel like Baire's category theorem could be somehow useful here... $\endgroup$
    – Lorenzo
    Sep 27, 2022 at 10:53
  • $\begingroup$ @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. $\endgroup$
    – Akira
    Sep 27, 2022 at 21:04
  • $\begingroup$ @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. $\endgroup$
    – Akira
    Sep 27, 2022 at 21:05
  • $\begingroup$ @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? $\endgroup$
    – Akira
    Sep 27, 2022 at 21:36
  • 1
    $\begingroup$ @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) $\endgroup$
    – Masacroso
    Sep 27, 2022 at 21:42

1 Answer 1


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.


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