When does $A\times B$ measurable imply both $A$ and $B$ measurable? Is Fubini's applicable? When $A\subset\mathbb R^n$ and $B\subset\mathbb R^m$ are Lebesgue measurable, then so is $A\times B\subset \mathbb R^{n+m}$ and $\mu(A\times B)=\mu(A)\cdot\mu(B)$. I'm being loose with the notation here assuming that we can understand that each $\mu$ is Lebesgue measure on the appropriate Euclidean space in each instance and ignoring any technical issues with specifying $\mathbb R^{n+m}$ vs $\mathbb R^{n}\times\mathbb R^{m}$.
I'm using the definition of Lebesgue outer measure:
$$\mu^*(E)=\inf\left\{\sum v(I_k)\mid E\subset \bigcup I_k, \text{ with closed boxes } I_k \right\}$$
and $\mu(E)=\mu^*(E)$ exists when for any $\epsilon>0$ there is an open set $G\supset E$ such that $\mu^*(G\setminus E)<\epsilon$. I'd like to avoid the Carathéodory's criterion
It is also true that nonmeasurable $A$ and measure zero $B$ gives $A\times B$ measurable with measure zero. Thus $A\times B$ measurable doesn't imply that $A$ and $B$ are both measurable. (Always meaning Lebesgue measurable etc.)
So I am wondering under what additional conditions, if any, do we have $A\times B$ measurable implying that $A$ and $B$ are both measurable? Is $\mu(A\times B)>0$ sufficient?
Will Fubini's theorem will give us something here?
Certainly $\mu(A\times B)=\mu^*(A\times B)=\mu^*(A)\cdot \mu^*(B)$. We also have that $\mathbf 1_{A\times B}$ the characteristic function on a measurable set and is thus measurable, and in fact $\mathbf 1_{A\times B}(x,y)=\mathbf 1_{A}(x)\cdot\mathbf 1_{B}(y)$. Thus we get
$$\int_{\mathbb R^{n+m}}\mathbf 1_A(x)\mathbf 1_B(y)=\int_{\mathbb R^{n+m}}\mathbf 1_{A\times B}(x,y)=\mu(A\times B)>0$$
But I don't quite see how I can say anything about $A$ and $B$ or $\mathbf 1_A$ and $\mathbf 1_B$ individually. If I try to use Fubini's to iterate the integral, I end up with $\mu(A)\mu(B)$. Thus if Fubini's is true then it makes mean think that $A$ and $B$ have to both be measurable when $\mu(A\times B)>0$. I don't see how Fubini's can be applied to $\mathbf 1_{A\times B}$ if we don't have specific knowledge about the measurability of $A$ and $B$.
Can I just say that
$$0<\mu(A\times B)=\int_{\mathbb R^{n}}\mathbf 1_A(x) \cdot \int_{\mathbb R^{m}}\mathbf 1_B(y)$$
and then I get to conclude that the right side must be the product of two positive numbers and hence $A$ and $B$ must both be measurable? That just feels a bit uncomfortable...
Now I'm not even sure if Fubini's is even relavant here.
When one set is nonmeasurable:
If $A\subset\mathbb R^n$ is nonmeasurable and $B\subset\mathbb R^m$ is measure zero, then I can almost work it out with Fubini's but not quite. I do believe we trivially get that $\mu(A\times B)=0$ though without much work.
$$\begin{aligned}
\int_{y\in\mathbb R^m}\mathbf 1_{A\times B}(x,y)&=
\begin{cases}
\mu(B) &\text{ if } x\in A\\
0 &\text{ if } x\not\in A\\
\end{cases}\\
&=0\cdot 1_{A}(x)=0\end{aligned}$$
Thus giving $\int_{\mathbb R^{n+m}}\mathbf 1_{A\times B} =\int_{\mathbb R^{n}}\int_{\mathbb R^{m}} \mathbf 1_{A\times B}=\int_{\mathbb R^{n}}0=0$ which is what I would hope for.
If we iterate the integral in the other order, then I run into issues though since
$$
\int_{x\in\mathbb R^n}\mathbf 1_{A\times B}(x,y) = \mathbf 1_{B}(y) \cdot \int_{x\in\mathbb R^n}\mathbf 1_{A}(x)$$
But $\mu(A)$ is undefined! For the purpose of this problem though, I can define $\int_{x\in\mathbb R^n}\mathbf 1_{A}(x)$ to be whatever I want though giving zero for the final result, but that kinda bothers me. I think maybe Fubini's theorem doesn't apply here since it starts off with the product measure?
I am sure there is probably some technical matter I am overlooking or making some mistake that is easy to fix.
 A: Given any set $H \in  \mathbb R^{n+m}$, any $x \in \mathbb R^n$ and any $y \in \mathbb R^m$ we define $H_x=\{y \in \mathbb R^m : (x,y) \in H\}$ and $H^y= \{x \in \mathbb R^n: (x,y) \in H\}$.

Let $A\subset\mathbb R^n$ and $B\subset\mathbb R^m$. If $A\times B$ is a Lebesgue measurable set in $\mathbb R^{n+m}$ and $\lambda_{n+m}(A\times B)>0$ then  $A$ and $B$ are both Lebesgue measurable.

Proof:
Since  $A\times B$ is Lebesgue measurable, there are $C, D$ a Borel measurable sets in  $\mathbb R^{n+m}$ and a set $N$ in  $\mathbb R^{n+m}$, such that:

*

*$\lambda_{n+m}(D)=0$;

*$N \subset D$;

*$A\times B = C \cup N$
Given any $x \in A$ , we have
$$ B = (A\times B)_x= C_x \cup N_x \tag{1}$$
We know that $C_x$ and $D_x$ are a Borel measurable sets in $\mathbb R^m$ and $N_x \subset D_x$.
Now, since  $\lambda_{n+m}(D)=0$, we have, Fubini's Theorem for product $\sigma$-algebra (in our case, Borel $\sigma$-algebra) that
$$ 0= \lambda_{n+m}(D) = \int_{ \mathbb R^{n+m}} 1_D d\lambda_{n+m} = \int_{ \mathbb R^{n}}  \left (\int_{ \mathbb R^{m}}  1_D d\lambda_{m} \right) d\lambda_{n} = \int_{ \mathbb R^{n}}  \lambda_{m}(D_x) d\lambda_{n}$$
So, for almost all $x \in \mathbb R^{n}$, $ \lambda_{m}(D_x) =0$. So, for almost all $x \in \mathbb R^{n}$, $C_x \cup N_x$ is Lebesgue measurable in $\mathbb R^{m}$.
Now, since  $\lambda_{n+m}(A\times B)>0$, we have that $\lambda_{n}^*(A)>0$ (otherwise, we would have  $\lambda_{n}(A)=0$ and $\lambda_{n+m}(A\times B)=0$). Since $\lambda_{n}^*(A)>0$, then there is at least one $x \in \mathbb R^{n}$ such that $C_x \cup N_x$ is Lebesgue measurable in $\mathbb R^{m}$. From $(1)$ we get that $B$ is Lebesgue measurable in $\mathbb R^{m}$.
In a similar way, we prove that $A$ is  Lebesgue measurable in $\mathbb R^{n}$.$\square$
Remark 1: The proof above can be easily adapted to prove the more general result:

If $H$ is a Lebesgue measurable set in $\mathbb R^{n+m}$, then, for almost all $x \in \mathbb R^{n}$ and  almost all $y \in \mathbb R^{m}$,  $H_x$ and $H^y$ are Lebesgue measurable.

Proof:
Since  $H$ is Lebesgue measurable, there are $C, D$ a Borel measurable sets in  $\mathbb R^{n+m}$ and a set $N$ in  $\mathbb R^{n+m}$, such that:

*

*$\lambda_{n+m}(D)=0$;

*$N \subset D$;

*$H = C \cup N$
Given any $x \in A$ , we have
$$ H_x= C_x \cup N_x $$
We know that $C_x$ and $D_x$ are a Borel measurable sets in $\mathbb R^m$ and $N_x \subset D_x$.
Now, since  $\lambda_{n+m}(D)=0$, we have, Fubini's Theorem for product $\sigma$-algebra (in our case, Borel $\sigma$-algebra) that
$$ 0= \lambda_{n+m}(D) = \int_{ \mathbb R^{n+m}} 1_D d\lambda_{n+m} = \int_{ \mathbb R^{n}}  \left (\int_{ \mathbb R^{m}}  1_D d\lambda_{m} \right) d\lambda_{n} = \int_{ \mathbb R^{n}}  \lambda_{m}(D_x) d\lambda_{n}$$
So, for almost all $x \in \mathbb R^{n}$, $ \lambda_{m}(D_x) =0$. So, for almost all $x \in \mathbb R^{n}$, $H_x = C_x \cup N_x$ is Lebesgue measurable in $\mathbb R^{m}$.
In a similar way, we prove that, for almost all $y \in \mathbb R^{m}$, $H_y$ is  Lebesgue measurable in $\mathbb R^{n}$.$\square$
Remark 2: If $A\subset\mathbb R^n$ is any set  and $B\subset\mathbb R^m$ has measure zero, then we have $A\times B \subset \mathbb R^n \times B$ and
$\lambda_{n+m}(\mathbb R^n\times B)=0$, so $\lambda_{n+m}(A\times B)=0$.
This shows that the additional condition $\lambda_{n+m}(A\times B)>0$ is actually required to allow the conclusion that $A$ and $B$ are both Lebesgue measurable.
A: This is a comment but it is too long:
With some little effort, you can actually see the following:
Suppose  $(X,\mathscr{F})$ and $(Y,\mathscr{G})$ are measurable spaces and $\mu^*$ and $\nu^*$ are outer measures such that

*

*The measurable sets obtained (with respect to Caratheodoty's cut condition) with respect to $\mu^*$ and  $\nu^*$ contained $\mathscr{F}$ and$\mathscr{G}$ respectively

*$A\subset X\times Y$ is measurable with respect to the outer measure
$(\mu^*\otimes\nu^*)(A)=\int\{\sum^n_{k=1}\mu(I_k)\nu(J_k): A\subset\bigcup^n_{j=1}I_j\times J_j\}$ where the $I_j$, $J_j$ are measurable (with respect to the Caratheodoty cut condition on for $\mu^*$ and $\nu^*$ respectively), and $A$ is $\sigma$--finite with respect to $\mu^*\otimes\nu^*$, then
$$(\mu^*\times\nu^*)(A)=\int\Big(\int \mathbb{1}_A\,d\nu^*)d\mu^*=\int\Big(\int \mathbb{1}_A\,d\mu^*)d\nu^*$$
that is
$$(\mu^*\times\nu^*)(A)=\int\nu^*(A_x)\,\mu^*(dx)=\int\mu^*(A^y)\,nu^*(dy)$$

*More generally, suppose $f:X\times Y\rightarrow\mathbb{R}$ is measurable with respect to $\mu^*\times\nu^*$ (in the sense that it is the limit of simple functions with steps in the $\sigma$--algebra that the outer measure $\mu^*\otimes\nu^*$ generates on $X\times Y$, and which contains $\mathscr{F}\times\mathscr{G}$). Then, Fubini-Tonelli's theorem states that $f$ is integrable with respect to $\mu^*\times\nu^*$ if and only if  $\{(x,y):f(x,y)\neq0\}$ is $\sigma$--finte w.r.t. $\mu^*\times\nu^*$, and
either of the interated integrals $\int_X\Big(\int_Y|f(x,y)|\nu^*(dy)\Big)\mu^*(dx)$ to $\int_Y\Big(\int_X|f(x,y)|\mu^*(dx)\Big)\nu^*(dy)$ exists and is finite. When that happened then you have
$$\int_{X\times Y}f(x,y)\mu^*\otimes\nu^*(dx,dy)=\int_Y\Big(\int_Xf(x,y)\mu^*(dx)\Big)\nu^*(dy)=\int_X\Big(\int_Yf(x,y)\nu^*(dy)\Big)\mu^*(dx)$$
(Notice the use of the outer measures)

In the case of Lebesgue measure in $\mathbb{R}^2$, one does not need to worry about $\sigma$--finiteness since almost by design, $\lambda\otimes\lambda$ is $\sigma$--finite. Here, the measurable sets with respect to Caratheodory's cut condition are what we call Lebesgue measurable sets, and this collection contains the Borel sets. In your particular case $f=\mathbb{1}_A\times\mathbb{1}_B$, where $A$ is not Lebesgue measurable (as a set in $\mathbb{R}$), and  $B$ is a set of Lebesgue measure $0$ (in $\mathbb{R})$), then $\mathbb{1}_{A\times B}$ has $(\lambda^*\otimes\lambda^*)$--measure $0$. This for example can be seen by taking
$f_n=\mathbb{1}_{A\times B}\mathbb{1}_{[-n,n]\times[-n,n]}$. this charactetistic funciton has measure zero since $f_n\leq \mathbb{1}_{[-n,n]}(x)\mathbb{1}_{B\cap[-n,n]}(y)$. As you know, the sets of outer measure $\lambda^*\times\lambda^*$-zero are Caratheodory measurable, and so their countable unions; hence $\mathbb{1}_{A\times B}$ is a set of outer measure $0$.
I hope this clarifies a few things. Of course you have to do some work to convince yourself that (1) and (2) hold, but it is not that difficult. You can also review again the procedures of Caratheodory to clearly understand what is going one with this business of outer measures.
