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I have to prove the following:

Let $f: \mathbb{R^2}\to \mathbb{R}$ such that $f_x:y\to f(x,y)$ is Borel measurable for all $x\in\mathbb{R}$ and that $f^y:x\to f(x,y)$ is continuous for all $y\in\mathbb{R}$. Prove that $f$ is Borel measurable.

What I have tried to do is to find a sequence of functions $f_n(x,y)$ s.t for a fixed $y$ $f_n(.,y)$ is a linear approximation of $f(.,y)$..

aduh
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Kelsey Middleton
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  • Here https://math.stackexchange.com/a/661135/266435 – Koro Oct 30 '22 at 17:50
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    @Koro: There is a general result by Caratheodory that deals with this kinds of situations and is well known in economics. I added a wiki answer in case you are interested in taking a look at the result. You may just assume (in the notation of my posting) that $X$ is a separable metric space and that $Y$ is any metric space. – Mittens Oct 30 '22 at 18:33

2 Answers2

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By the continuity of $f^y$ we have $$f(x,y) = \lim_{n \to \infty}f(\lfloor nx \rfloor / n, y).$$ By the measurability of $f^x$, we see that $f$ is the pointwise limit of a sequence of Borel measurable functions, and hence is itself Borel measurable.

aduh
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    how is $f(\lfloor nx \rfloor/n, y)$ borel? – Airdish May 19 '19 at 23:08
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    @Airdish: The map $\phi:(r, x)\mapsto f(r, x)$ on $\mathbb{Q}\times \mathbb{R}$ meaurble since ${(r, x):f(x, r)<a}=\bigcup_{r\in\mathbb{Q}}{r}\times (f_r)^{-1}(-\infty,a)$. – Mittens Oct 30 '22 at 19:08
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There is a general result that can be applied in this situation:

Definition C: Suppose $(S,\mathscr{S})$ is a measurable space, and $(X,\tau_X)$ and $(Y,\tau_Y)$ topological spaces equipped with the Borel $\sigma$-algebras $\mathscr{B}(\tau_X)$ and $\mathscr{B}(\tau_Y)$. A function $f:S\times X\rightarrow Y$ is called a Caratheodory function if

  1. $f_s:x\mapsto f(s,x)$ is continuous for each $s\in S$,
  2. $f_x:s\mapsto f(s,x)$ is $\mathscr{S}-\mathscr{B}(\tau_Y)$ measurable.

Here is a well known result by Carathéodory.

Theorem: If $f$ is as in definition C, $(X,\tau_X)$ is metrizable and separable and $(Y,\tau_Y)$ is metrizable, then $f$ is $\mathscr{S}\otimes\mathscr{B}(\tau_X)-\mathscr{B}(\tau_Y)$ measurable.

A proof of this can be found in Aliprantis, C. and Border, K., Infinite dimensional analysis: A hitchhiker's guide., Third Edition, Springer. pp 153.

In the context of the OP, $S=\mathbb{R}$, $\mathscr{S}=\mathscr{B}(\mathbb{R})$, $X=Y=\mathbb{R}$ also equipped with the Borel $\sigma$-algebra.

Mittens
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