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Suppose $(X,\mathcal {M})$ is a measurable space. For function $f(r,x): \mathbb R\times X\to\mathbb R$, suppose each r-section of it is $\mathcal M$-measurable. For constant $R\in\mathbb R$, do we have $\limsup \limits_{r\to R}f(r,x)$(this is a function of $x\in X$) always $\mathcal M$-measurable? The definition of $\limsup \limits_{r\to R} f(r)$ is $\inf\limits_{\delta>0}\{\sup\limits_{0<|r-R|<\delta}f(r)\}$. Thanks!

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Could you please refer me to the text containing this statement? I tried to prove it but failed. –  zzzhhh Mar 9 '11 at 15:45
    
I'm sorry - I misread the question at first (the Firefox 4 MathML display problem). The version that is usually proved is for a sequence of measurable functions, rather than a family parametrized by the reals. Since you only assumed that each section is measurable, I'm not sure that the statement is even true. –  Carl Mummert Mar 9 '11 at 16:01
    
Yeah, the latex formula display is really problematic on this website. I simulated the proof of the sequence version and express $\left(\sup \limits_{0<|r - R| < \delta } f\right)^{-1}((a,\infty])$ as $\bigcup\limits_{0 <|r-R|<\delta}{f^{-1}((a,\infty ])}$. But the union on the right is not countable, so the result is not necessarily measurable. This is where I failed to prove it. I tried to restrict r to be rational numbers only, but seems not work. –  zzzhhh Mar 9 '11 at 16:13
    
The question I have is whether the statement would become provable if you assumed that the entire function $f(r,x)$ is measurable, not just that each section is measurable. –  Carl Mummert Mar 9 '11 at 16:18
    
@Carl That is a good question. Could I suggest that you ask it separately? –  Byron Schmuland Mar 9 '11 at 16:53

1 Answer 1

up vote 3 down vote accepted

The principle is disprovable in ZFC. (I had a previous answer where I showed that it was disprovable from ZFC+CH, which I've edited with an improved solution).

First, fix some map $g$ from $(0,1)$ to itself such that for every $s \in (0,1)$ there are $r$ arbitrarily close to $0$ such that $g(r) = s$. The map $g$ can be taken to be $(1+\sin(1/r))/2$ but the exact definition isn't important.

Let $A$ be a nonmeasurable subset of $(0,1)$. For $r \in (0,1)$ we define the function $f_r: (0,1) \to \{0,1\}$ as the characteristic function of $A \cap \{g(r)\}$. So $f_r(g(r))$ is $1$ if $g(r) \in A$, and $f_r(g(r)) = 0$ otherwise, and $f_r(x) = 0$ for all $x \not = g(r)$.

Let $f: (0,1) \to \{0,1\}$ be the limit function $f(s) = \lim_{r \to 0^+} \sup_{\{t: 0 < t < r\}} f_t(s)$. Then $f$ is the characteristic function of $A$, and so $f$ is not measurable.

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I get it. This an ingenious counterexample. Thank you very much, professor! –  zzzhhh Mar 10 '11 at 7:17
    
And if it happens that f is continuous for each of its r-section, we can have the desired result, because the union of any collection of open sets is still open. –  zzzhhh Mar 10 '11 at 8:00

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