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Consider two measure spaces $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$, where $\mu\times\nu(X\times Y)>0$. Given two measurable functions $f:X\to \mathbb{R}$ and $g:Y\to\mathbb{R}$ such that

$$f(x) = g(y) \qquad\mu\times\nu \,\,\text{a.e,}$$

does it follow that there exists a constant $\lambda$ such that $f(x)=\lambda$ for $\mu$-a.e. $x$ and $g(y)=\lambda$ for $\nu$-a.e. $y$?

Clarification: The displayed statement means that $\mu\times\nu\big(\{(x,y):f(x)\neq g(y)\}\big)=0$.

I know that this is true when you assume $\mu$ and $\nu$ are $\sigma$-finite, by applying Fubini's theorem to $\int |f(x)-g(y)|\,d(\mu\times\nu)$. In the general case, the only first step I can think of is to find a countable set of rectangles with arbitrarily small total area where $f(x)=g(y)$ on the complement of their union. But there I'm stuck.

Mike Earnest
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  • I thought it might be (generally) true that $\int_{X\times Y} h(x,y)d(\mu\times \nu) \ge \int_X \int_Y h(x,y)d\nu(y)d\mu(x)$ for any measurable $h$ on $X\times Y$ (which would solve your problem), but if you take $X=Y=[0,1]$, $d\mu = dx$ = Lebesgue, $\nu = $ counting, and $h(x,y) = 1_{x = y}$, then $\int_{X\times Y} h(x,y)dxd\nu(y) = 0$ while $\int_X\int_Y h(x,y)d\nu(y)dx = 1$. Can you confirm this is correct, if you have the time? Thanks. – mathworker21 Jan 04 '20 at 03:40

2 Answers2

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Ok, after a year of struggling with this I finally think I have an answer! Figured I should post for whoever else is curious: it does follows that $f,g=\lambda$ a.e. for some $\lambda$.

First off, without loss of generality we can assume $f(x),g(y)\in[0,1]$ for all $x,y$. If not, just compose them with $h(s)=\frac{\tan^{-1}(s)+\pi/2}{\pi}$; since $h$ is injective, $h\circ f=$ constant a.e. implies $f=$ constant a.e.

Let $$B_{n,b}=\{x\in[0,1]:\text{the nth binary digit of x is } b\},$$ for $b=0,1$, $n\ge1$. Let $F_{n,b} = f^{-1}(B_{n,b})$ and $G_{n,b} = g^{-1}(B_{n,b})$.

Claim: For each $n$, we have either $F_{n,0}\times G_{n,0}$ is full measure, or $F_{n,1}\times G_{n,1}$ is full measure. To see this, first note that both $\mu\times\nu(F_{n,0}\times G_{n,1})$ and $\mu\times\nu(F_{n,1}\times G_{n,0})$ are zero since they are subsets of $\{f(x)\neq g(y)\}$. These, combined with $F_{n,0}+F_{n,1}=\mu(X)>0$ and $G_{n,0}+G_{n,1}>0$, show that $$ \mu(F_{n,0})>0\implies \nu(G_{n,1})=0\implies \nu(G_{n,0})>0\implies \mu(F_{n,1})=0 $$ and $$ \mu(F_{n,0})=0\implies \mu(F_{n,1})>0\implies \nu(G_{n,0})=0. $$ Thus, either both $F_{n,0}$ and $G_{n,0}$ are full measure, or both $F_{n,1}$ and $G_{n,1}$ are, proving my claim.

Let $b_n$ be the bit for which $F_{n,b_n}\times G_{n,b_n}$ is full measure. Then $$\bigcap_{n\ge 1}F_{n,b_n}\times G_{n,b_n}=\left(\bigcap_{n\ge 1}F_{n,b_n}\right)\times \left(\bigcap_{n\ge1}G_{n,b_n}\right)$$ will have full measure as well, implying both $\bigcap_{n\ge 1}F_{n,b_n}$ and $\bigcap_{n\ge 1}G_{n,b_n}$ are full measure. But these are precisely $f^{-1}(\lambda)$ and $g^{-1}(\lambda)$, where the $n^{th}$ binary digit of $\lambda$ is $b_n$. Thus, both $f^{-1}(\lambda)$ and $g^{-1}(\lambda)$ have full measure, so $f(x)=\lambda$ a.e. and $g(y) = \lambda$ a.e, with respect to $\mu,\nu$.

If this proof is wrong/needs clarification, please comment!

Mike Earnest
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  • Looks completely correct to me. (I had an idea about dyadic intervals on $\mathbb{R}$, but I don't think it works anymore). – Matt Rigby Sep 21 '14 at 22:51
  • In fact, it looks like the dyadic interval idea I had only works if it's bounded to make sure you only start with two intervals; in which case, a linear transformation changes it to being on $[0,1]$ and it becomes exactly what you have done due to the binary expansions being precisely the same as writing the number as an intersection of dyadic intervals! – Matt Rigby Sep 21 '14 at 23:03
  • Question: How do we know that each $B_{n,b}$ is actually Borel measurable, or does it not matter? – user360187 Aug 10 '16 at 03:58
  • It matters! The sets $B_{n,b}$ are measurable since they are finite unions of half open intervals e.g. $B_{1,0}=[0,\tfrac12)$, $B_{2,1}=[\tfrac14,\frac12)\cup[\tfrac34,1)$, $B_{3,0}=[0,\tfrac18)\cup[\tfrac14,\tfrac38)\cup[\tfrac12,\tfrac58)\cup[\tfrac34,\tfrac78)$. – Mike Earnest Aug 10 '16 at 04:26
  • Alternatively, we can define the "nth bit of x function" by $f_n(x)=1$ if $\lfloor 2^n x\rfloor$ is odd and $0$ otherwise. Then $B_{n,b}=f_{n}^{-1}({b})$, so you just need to prove $f_n$ is measurable. – Mike Earnest Aug 10 '16 at 04:33
  • Ah right thanks Mike – user360187 Aug 10 '16 at 05:35
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I have a counterexample.

Let $X=Y=\lbrace \ast,\#\rbrace$ with $\mathcal M=\mathcal N=\wp(X)$. Define $\mu$ on $\mathcal M$ by setting $\mu(\ast)=0$ and $\mu(\#)=\infty$ and $\nu$ by switching the roles of the two values taken by $\mu$, i.e. $\nu(\ast)=\infty$ and $\nu(\#)=0$. Let $f=g\colon X=Y\longrightarrow \mathbb R$ be the measurable function defined by $f(\ast)=0$ and $f(\#)=1$.

We have $f=1$ $\mu$-a.e. and $g=0$ $\nu$-a.e.

Vincenzo Zaccaro
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  • But this does not satisfy the assumption $f(x)=g(y)\quad \mu× \nu$ a.e. The set of ordered pairs $(x,y)$ where $x\neq y$ has a product measure equal to $\infty$. – Mike Earnest Aug 03 '20 at 17:31