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Suppose $\{ (G_i, \mathcal{G}_i), i=1,2 \}$ are measurable spaces. Their product measurable space is $(\prod_{i=1}^2 G_i, \prod_{i=1}^2 \mathcal{G}_i)$.

$(F, \mathcal{F})$ is another measurable space.

$g:\prod_{i=1}^2 G_i \rightarrow F $ is a mapping. Define sections of $g$ on each component space $G_i$ as follows

$\forall x_1 \in G_1$ define $g_{x_1}: G_2 \rightarrow F$ as $g_{x_1}(x_2) := g(x_1,x_2)$;

$\forall x_2 \in G_2$ define $g_{x_2}: G_1 \rightarrow F$ as $g_{x_2}(x_1) := g(x_1,x_2)$.

  1. Similar to Theorem 18.1 of Probability and Measure by Billingsley, it can be shown that if $g$ is measurable, then its sections are all measurable. I was wondering if the converse is also true, i.e. if its sections are all measurable, is $g$ also measurable?
  2. If it is true, will it still be true for arbitrarily many $\{ (G_i, \mathcal{G}_i), i=I \}$?
  3. I am also wondering about the same questions when replacing measurable spaces with topological spaces, and measurability of a mapping by continuity of a mapping? (I will move this topological case to another post if the answers are not essentially based on the same idea.)

Thanks and regards!

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In 1920 Sierpiński constructed an example of a non-Lebesgue measurable subset $S$ of $\mathbb{R}^2$ intersecting each line in at most two points. This implies of course that $S$ intersects each horizontal and vertical line in at most two points, so that the sections of the characteristic function $[S]: \mathbb{R}^2 \to \mathbb{R}$ are all measurable. –  t.b. Jan 4 '12 at 2:43

3 Answers 3

up vote 3 down vote accepted

Take for example $G_1 = G_2 = [0,1]$ with the Borel $\sigma$-algebra, and let $A$ be a non-Borel subset of the diagonal. (For example, let $E$ be a non-Borel subset of $[0,1]$ and let $A = \{ (x,x) : x \in E \}$.) Then the indicator function $1_A$ is not measurable, but you can check that all of its sections are.

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I was wondering why when $E$ is non-Borel, the intersection of $E$ and the diagonal is also non-Borel? –  Tim Feb 14 '13 at 14:21
    
@Tim: Read again; $A$ is not the intersection of $E$ with the diagonal. $E$ is a subset of $[0,1]$, not $[0,1]^2$. –  Nate Eldredge Feb 14 '13 at 17:20
    
Thanks. Sorry,for misunderstanding. Why is $A$ non-Borel measurable? –  Tim Feb 14 '13 at 17:38
    
@Tim: Consider the map $f : [0,1] \to [0,1]^2$ defined by $f(x) = (x,x)$. $f$ is continuous hence Borel measurable. But $E = f^{-1}(A)$, so if $A$ were Borel $E$ would also be Borel. –  Nate Eldredge Feb 14 '13 at 20:19

For (3) consider the function $$f:\mathbb{R}^2\to\mathbb{R}:\langle x,y\rangle\mapsto \begin{cases}\frac{2xy}{x^2+y^2},&\text{if }\langle x,y\rangle\ne\langle 0,0\rangle\\\\ 0,&\text{if }x=y=0\;; \end{cases}$$ $f$ is easily seen to be discontinuous at the origin, but all of its sections are continuous.

Added: For background and references on separate versus joint continuity see Z. Piotrowski, Separate versus joint continuity - an update, Proc. 29th Spring Conference Union Bulg. Math. Lovetch (2000), 93-106, available here as Nr. 51.

T.O. Banakh, O.V. Maslyuchenko, & V.V. Mykhaylyuk show in Discontinuous separately continuous functions and near coherence of $P$-filters that for every non-discrete Tikhonov space $X$ there are a Tikhonov space $Y$ with a unique non-isolated point and a bounded separately continuous $f:X\times Y\to\mathbb{R}$ that is not jointly continuous and use this result to show

Theorem: For any non-discrete Tikhonov space $X$ there are a $\sigma$-bounded Abelian topological group $G$ and separately continuous $h:X\times G\to \mathbb{R}$ that is not jointly continuous.

The paper is available here as Nr. 125.

For (1), Maxim R. Burke, Borel measurability of separately continuous functions, Topology and its Applications 129 (2003), 29–65, mentions that Walter Rudin, in Lebesgue’s first theorem (in L. Nachbin (Ed.), Mathematical Analysis and Applications, Part B, in Advances in Mathematics Supplementary Studies, Vol. 7B, Academic Press, New York, 1981, pp. 741–747), gives a ‘very simple example of a non-Borel separately continuous function $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}^\mathbb{R}$’, but I’ve not seen it. In Borel measurability of separately continuous functions, II, Topology and its Applications 134 (2003), 159–188, he notes that D.K. Burke & R. Pol, On Borel sets in function spaces with the weak topology, J. London Math. Soc.(2) 68 (2003), 725–738, have shown that if $X$ is an infinite compact $F$-space without isolated points or a Baire $P$-space without isolated points, then the evaluation map $e:X\times C_p(X)\to\mathbb{R}:\langle x,f\rangle\mapsto f(x)$ is not Borel measurable, though it is separately continuous. (A space $X$ is an $F$-space if every cozero set in $X$ is $C^*$-embedded in $X$. there are many compact $F$-spaces, since $\beta X\setminus X$ is an $F$-space whenever $X$ is a $\sigma$-compact, locally compact space. A $P$-space is a topological space in which every $G_\delta$-set is open.)

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+1 Thanks! Especially the references are great, and you even mentioned the mixed case of jointly measurability and separate continuity that I didn't think of. If I could, I wish I could have accepted all three replies. My bad that I asked questions in two different areas. –  Tim Jan 4 '12 at 21:14

For continuity, the answer is no (at least if I'm interpreting your definition of continuity for "sections" correctly). Here is the standard counterexample. Define a function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ by $f(x,y)=\frac{xy}{x^2+y^2}$ if $(x,y) \neq (0,0)$ and $f(0,0)=0$. Clearly $f$ is continuous away from $(0,0)$, but it is an easy exercise (that I usually give when teaching calculus) that $f$ is not continuous at $(0,0)$. However, the functions $f(0,y)=0$ and $f(x,0)=0$ are definitely continuous.

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