Let $f:X\to Y$ is a measurable function. Banach indicatrix $$ N(y,f) = \#\{x\in X \mid f(x) = y\} $$ is the number of the pre-images of $y$ under $f$. If there are infinitely many pre-images then $N(y,f) = \infty$.

Let $X\subset\mathbb R^n$, $Y\subset\mathbb R^m$ with Lebesgue measure.

I am interested to know if $N(y,f)$ is a measurable function (?)

Let $X$ be a separable metric space and let $f(A)$ be $\mu$-measurable for all Borel subsets $A$ of $X$. Let $\zeta(S) = \mu(f(S))$ for $S\subset X$ and let $\psi$ be the measure on $X$ defined by the Carathéodory construction from $\zeta$. Then $$ \psi(A) = \int\limits_{A}N(y,f)\, d\mu_{Y} $$ for every Borel set $A\subset X$.

Does it say anything about measurability of $N(y,f)$ ?

  • $\begingroup$ You need to be more specific about what spaces $X,Y$ you want to consider. In this generality, there are trivial counterexamples (take the $\sigma$-algebra $\emptyset , Y$ on $Y$). $\endgroup$ May 16, 2015 at 17:00
  • $\begingroup$ @ChristianRemling Agree, I am thinking about Lebesgue measure. $\endgroup$ May 18, 2015 at 4:32
  • $\begingroup$ You may find it helpful to know that if $X$ is a Borel set and $f$ is Borel measurable, then for any Borel set $A \subset X$, the set $f(A)$ is analytic and hence universally measurable, so the hypothesis of the theorem from Federer is satisfied. (Presumably part of the conclusion of that theorem is the statement that $N(y,f)$ is measurable, otherwise the integral makes no sense.) $\endgroup$ May 18, 2015 at 16:19
  • $\begingroup$ This is a special case of the co-area formula. To get the measurability of the quantity you are interested in it suffices to assume that $X$ and $Y$ are rectifiable, of the same dimension, and $f$ is locally Lipschitz. For a proof see Lemma 5.2.5 in the book Geometric Integration Theory by S. Krantz and H. Parks. $\endgroup$ Nov 5, 2015 at 11:00
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    $\begingroup$ @LiviuNicolaescu Yes it really is the co-area formula. However the aim was to weak the regularity of the mapping $f$. $\endgroup$ Nov 5, 2015 at 11:48

3 Answers 3


A counterexample.

Let $F : \mathbb R \to \mathbb R$ be the Cantor singular function. $F$ is continuous. Let $C \subseteq \mathbb R$ be the middle-thirds Cantor set. $C$ has Lebesgue measure zero. $F$ maps $C$ onto $Y = [0,1]$. Let $M \subseteq [0,1]$ be a non-measurable set. For simplicity, remove from $M$ any points with two different binary expansions. Then $X := F^{-1}(M)$ is a subset of $C$, so it is Lebesgue measurable. Let $f : X \to Y$ be the restriction of $F$. $f$ is continuous, hence measurable. Next: $$ \#\{x \in X : f(x)=y\} = \begin{cases} 0,\qquad y \not\in M \\ 1,\qquad y \in M \end{cases} $$ Note there are no points with $2$ or more pre-images, since those would be points with two different binary expansions, and we removed those.

Since $M$ is a non-measurable set, we have a counterexample.

Compare the two bullet-points in the question.
For the first: in this example $X$ is not an interval.
For the second: in this example the property "$A$ Borel (in $X$) implies $f(A)$ measurable" fails.


This is an attempt to rid of demand of continuity. The following proof is essentially an adaptation of Banach's original proof in case of continuous function defined on segment $[a,b]$. See also https://math.stackexchange.com/a/144832/23566.

Lemma Let $f\colon \mathbb R^n \to \mathbb R^m$ be a measurable function and an image $f(B)$ is measurable for any Borel measurable set $B$. Then the Banach Indicatrix $N(y,f)$ is measurable.

Proof. We use a dyadic decomposition. For each integer $k\geq 0$ consider collection of cubes $\{P^k_i\}$ of the form $$ P^k_i = (a_1^i\cdot2^{-k},(a_1^i+1)\cdot2^{-k}]\times\cdots\times(a_n^i\cdot2^{-k},(a^i_{n+1})\cdot2^{-k}], $$ where the $a^i_j$ are all integers. The properties we need are following: cubes $P^k_i$ are disjoint; $\mathbb R^n = \bigcup\limits_{i=1}^{\infty}P^k_i$; $\operatorname{diam} P^k_i = \sqrt{n}2^{-k}\to 0 $ as $k\to\infty$.

For $y \in \mathbb R^m$ and $i\in \mathbb N$ let $$ L_{i}^{(k)}(y) = \begin{cases} 1, & \text{if } y \in f(P_{i}^{(k)}), \\ 0, & \text{if } y \not\in f(P_{i}^{(k)}). \end{cases} $$ The functions $L_{i}^{(k)}(y)$ are non-negative and measurable because the set $f(P_{i}^{(k)})$ is measurable. Therefore the sum $$ N_k(y) =\sum\limits_{i=1}^{\infty}L_{i}^{(k)}(y) $$ is also measurable. Thus, the sequence $(N_k)_{k=1}^\infty$ of measurable functions is increasing and therefore the pointwise limit $$ N^*(y) = \lim_{k\to\infty} N_k(y) $$ exists and is a measurable function of $y$.

Note that $N_k(y)$ simply counts on how many of the cubes $P_{i}^{(k)}$ the function $f$ attains the value $y$ at least once. Thus $N(y,f) \geq N_k(y)$ for all $k$, so $N(y) \geq N^*(y)$.

Let us argue that $N^*(y) \geq N(y,f)$. Let $q$ be an integer such that $N(y,f) \geq q$. Then there exist $q$ different points $x_1,\dots,x_q$ such that $f(x_j) = y$. If $k$ is so large that points $x_1,\dots,x_q$ are in separated cubes $\{P_{i_j}^{(k)}\}_1^q$ then $N_k(y) \geq q$. This shows $N^*(y) \geq N(y,f)$ and thus $N^*(y) = N(y,f)$, establishing measurability.


After a while I came to the following

Theorem Let $f:X\to Y$ be a $\mu_X$-measurable mapping, and $A\subset X$ be a Borel set. Then $f$ can be redefined on a set of $\mu_X$-measure zero in such a way that the Banach indicatrix $N(y,f,A)$ is a $\mu_Y$-measurable function.

Partly this was known though I've wrote a proof here.

  • $\begingroup$ Very interesting. If you have any related results, please let me know. I have been thinking about similar problems recently. $\endgroup$ May 26, 2020 at 18:34

This is a corollary of Rokhlin's theorem on classification of measurable partitions in Lebesgue spaces from his famous paper "On the fundamental ideas of measure theory". The Russian original is available. The statement you need is (I) at the beginning of p.138. Alas, the way measure theory is still taught nowadays, most mathematical students are nor aware of Rokhlin's theory.

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    $\begingroup$ In light of Gerald Edgar's counterexample, and especially for non-readers of Russian, could you elaborate on the hypotheses which are needed for Rokhlin's theorem? $\endgroup$ May 18, 2015 at 16:16
  • $\begingroup$ Sure - Rokhlin's theorem refers to the measure category, i.e., everything is understood modulo subsets of measure 0, and the map is supposed to be measure class preserving. This was my interpretation of the question since the Lebesgue measure was mentioned several times. $\endgroup$
    – R W
    May 18, 2015 at 21:23
  • $\begingroup$ "Measure class preserving" means "maps null sets to null sets"? $\endgroup$ May 18, 2015 at 21:27
  • $\begingroup$ Yes, precisely that $\endgroup$
    – R W
    May 18, 2015 at 21:49

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