Is the following version of Fubini's theorem true? All the following integrals are with respect to Lebesgue measure.
Suppose $f(x,y):\mathbf{R}^n \times \mathbf{R}^m=\mathbf{R}^{n+m}\to \mathbf{R}\cup\{-\infty, \infty\}$ is Borel and integrable (or Borel and nonnegative). Then all the following integrals exist and
There exists Borel sets $A\subset\mathbf{R}^{n}$ and $B\subset\mathbf{R}^{m}$ of full Lebesgue measure, such that for all $x\in A$ and for all $y\in B$ functions $f_x(y)=f(x,y)$ and $f_y(x)=f(x,y)$ respectively, are Borel.
Functions $$\mathbf{R}^{n}\ni x\mapsto \int\limits_{\mathbf{R}^m} f_x(y) d\lambda_m(y) \in \mathbf{R}\cup\{-\infty, \infty\}$$ and $$ \mathbf{R}^{M}\ni y\mapsto \int\limits_{\mathbf{R}^N} f_y(x) d\lambda_m(x) \in \mathbf{R}\cup\{-\infty, \infty\}$$ (we put $0$ where respectively $f_x$ and $f_y$ aren't Borel), are Borel.
$$\begin{align*} \int\limits_{\mathbf{R}^{n+m}} f d\lambda_{n+m} &=\int\limits_{\mathbf{R}^n} \Big(\int\limits_{\mathbf{R}^m} f_x(y) d\lambda_m(y)\Big) \ d\lambda_n(x) \\ &= \int\limits_{\mathbf{R}^m} \Big(\int\limits_{\mathbf{R}^n} f_y(x) d\lambda_n(x)\Big) \ d\lambda_m(y) \end{align*}$$
If not, is there anything close to that true? Like: are the sets $A$ and $B$ just $\mathbf{R}^n$ and $\mathbf{R}^m$, respectively? Or we can't say that they are Borel, just Lebesgue measurable? Or maybe just two first points are true? Or is it required that values of $f$ are finite (which would be fine for me since i'd like to use the theorem to probability densities, which i assume to be Borel).
At Uni i only had version for Lebesgue measure and Lebesgue measurable functions. I'm not familiar with terms like $\sigma$-finite measure, complete measure or general product measure (i read definitions but haven't worked on them at all) and got confused with general statements of Fubini's theorem. What caught my eye is the notion of completness which might have a problem with being Borel.