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If $f\left(x,\cdot\right)$ is measurable for every $x$ and $f\left(\cdot,y\right)$ is measurable for every $y$, is $f$ necessarily measurable?

More precisely, let $\left(\Omega_i,\mathcal{A}_i\right)$ be measurable spaces, $i\in\left\{1,2,3\right\}$. Let $f:\Omega_1\times\Omega_2\rightarrow\Omega_3$. For every $\omega_1\in\Omega_1$, $\omega_2\in\Omega_2$ define

$$ f_1^{\left(\omega_1\right)}:\Omega_2\rightarrow\Omega_3,\space\space f_1^{\left(\omega_1\right)}\left(\omega_2\right):=f\left(\omega_1,\omega_2\right) $$

$$ f_2^{\left(\omega_2\right)}:\Omega_1\rightarrow\Omega_3,\space\space f_2^{\left(\omega_2\right)}\left(\omega_1\right):=f\left(\omega_2,\omega_1\right) $$

It is known (e.g. Schilling, Theorem 13.10 iii) that if $f$ is $\mathcal{A}_1\otimes\mathcal{A}_2/\mathcal{A}_3$-measurable, then $f_1^{\left(\omega_1\right)}$ is $\mathcal{A}_2/\mathcal{A}_3$-measurable for every $\omega_1\in\Omega_1$ and $f_2^{\left(\omega_2\right)}$ is $\mathcal{A}_1/\mathcal{A}_3$-measurable for every $\omega_2\in\Omega_2$.

But does the converse hold as well?

In comparison, both directions hold in the following, related result (Schilling, Theorem 13.10 ii): $f:\Omega_3\rightarrow\Omega_1\times\Omega_2$ is $\mathcal{A}_3/\mathcal{A}_1\otimes\mathcal{A}_2$-measurable iff $\pi_i\circ f$ is $\mathcal{A}_3/\mathcal{A}_i$-measurable ($i\in\left\{1,2\right\}$), with $$\pi_i:\Omega_1\times\Omega_2\rightarrow\Omega_i,\space\space \pi_i\left(\left(\omega_1,\omega_2\right)\right):=\omega_i$$

References

Schilling, René L. Measures, Integrals and Martingales. 2005

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  • 2
    $\begingroup$ if you take a function $f: \mathbb{R} \times \mathbb{R}\rightarrow \mathbb{R}$ to be the characteristic of the vitaly as a subset of $\mathbb{R} \times \mathbb{R}$ and rotated by $45 $ degrees? $\endgroup$ – clark Apr 12 '13 at 17:52
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Assuming CH... there is a set $E$ in the unit square $[0,1]\times[0,1]$ such that every vertical cross-section $$ \{y : (a,y) \in E\} $$ is countable, and every horizontal cross-section $$ \{x:(x,b) \in E\} $$ has countable complement. Then by Fibini's theorem, the set is not Borel measurable. Not even Lebesgue measurable.

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  • $\begingroup$ If you don't want to assume CH, make an example in $\omega_1 \times \omega_1$, but then you have to explain about Borel sets, and the non-fixed zero-one measure. $\endgroup$ – GEdgar Apr 12 '13 at 21:36
  • $\begingroup$ Thanks. An interesting way to approach the problem. Unfortunately, i'm a little weak on my set theory, so i'll stick with my/clark's answer. $\endgroup$ – Evan Aad Apr 12 '13 at 22:05
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No, the converse doesn't hold. The following counter-example is based on clark's comment to my original post.

Set $\Omega_i:=\mathbb{R}$, $\mathcal{A}_i:=\mathfrak{B}$ ($\mathfrak{B}$ being the standard Borel field on the real line). Let $C\subseteq\mathbb{R}$ be any non Borel set, e.g. the Vitali set. Let $T:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}\times\mathbb{R}$ be a rotation of the plane by an angle that is not a multiple of $\frac{\pi}{2}$. Define $C'$ to be the set $\left\{\left(x,0\right)\space:\mid\space x\in C\right\}$, i.e. the natural embedding of $C$ in the $2$-dimensional "$x$-axis".

$C'\notin\mathfrak{B}\otimes\mathfrak{B}$, since otherwise $C$ would be $\in\mathfrak{B}$, as a section of $C'$ (cf Halmos, p. 141), contrary to assumption. Define $D:=T\left(C'\right)$. Since $T$ is surjective and measurable (it is measurable since it is linear), $D\notin\mathfrak{B}\otimes\mathfrak{B}$. Define $f:=\mathbb{1}_D$. Thus $f$ is not $\mathfrak{B}\otimes\mathfrak{B}/\mathfrak{B}$-measurable. Since $T$ is linear and $C'$ lies on a line through the origin, so does $D$, but since by definion of $T$ this line is perpendicular to neither "axis", $D$'s sections consist of at most a single point, so for all $x\in\mathbb{R}$, $f_1^{(x)}$ is $\mathfrak{B}/\mathfrak{B}$-measurable and likewise $f_2^{(y)}$ for all $y\in\mathbb{R}$.

References

Halmos, Paul Richard. Measure Theory. 1974

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There is a bijection from reals to reals whose graph has full outer measure so that the characteristic function of the graph of this bijection is a counterexample.

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  • $\begingroup$ Thanks, but i find your answer a little obscure. I've provided another answer based on clark's comment to my original post. Thank you regardless. $\endgroup$ – Evan Aad Apr 12 '13 at 20:58
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    $\begingroup$ I interpreted "f(x, y)" is measurable as "pre-imgaes of Borel sets are Lebesuge measurable". $\endgroup$ – hot_queen Apr 13 '13 at 17:14

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