1. a question about Lebesgue measurable function

Please give a Lebesgue measurable subset E of Rn
and a Lebesgue measurable and finite-valued function f on E
satisfies
there doesn't exist a sequence of Lebesgue measurable subsets {Ek}k=1,2,...
of E
such that
(1)the untion of Ek is E;
(2)the restriction of f to every set Ek is continuous on Ek.

2. Re: a question about Lebesgue measurable function

I think the function that is 1 if every coordinate is rational and 0 elsewhere might do what you want.

Edit: Oops, no it doesn't. Obviously.

Maybe divide R^n into equivalence classes where x~y iff x-y has all rational coordinates. Well-order the equivalence classes and set your function equal to 2^{-n} on the first countable infinity of them, and zero elsewhere.

3. Re: a question about Lebesgue measurable function

If the range of function f is countable,then f certainly doesn't satisfy.

4. Re: a question about Lebesgue measurable function

Yeah, you're right, else we'd just get the sequence as inverse images of points in the range. Okay, there's nothing keeping us from assigning a different value to each equivalence class. There is an injection from the equivalence classes into [0,1] for example.

I may be overthinking this. I think the key point is going to be that a complete metric space cannot be written as a countable union of nowhere dense sets. So one of our subsets is dense on some open ball in R^n. We need a function so ugly that its restriction to any dense subset of any open ball cannot be continuous. I thought along this line for a while but couldn't get further. Maybe it's not the way to go, or maybe my analysis is too rusty. Good luck!

5. Re: a question about Lebesgue measurable function

Originally Posted by Tinyboss
Maybe divide R^n into equivalence classes where x~y iff x-y has all rational coordinates. Well-order the equivalence classes and set your function equal to 2^{-n} on the first countable infinity of them, and zero elsewhere.
I don't think this set is measurable since it's the product of copies of a non-measurable one.

Edit: As for the question, for example if $\displaystyle E\subset \mathbb{R}^n$ is compact then by Lusin's theorem we can obtain sets with the desired properties with the exception that $\displaystyle \cup E_k = E\setminus Z$ with $\displaystyle \lambda (Z)=0$, so maybe we could construct a function for which this cannot be improved.

6. Re: a question about Lebesgue measurable function

Originally Posted by mathabc
Please give a Lebesgue measurable subset E of Rn
and a Lebesgue measurable and finite-valued function f on E
satisfies
there doesn't exist a sequence of Lebesgue measurable subsets {Ek}k=1,2,...
of E
such that
(1)the untion of Ek is E;
(2)the restriction of f to every set Ek is continuous on Ek.
The way I read this it's impossible. Let $\displaystyle f:E\to\mathbb{R}^n$ be a finite-valued Lebesgue measurable function. The codomain of $\displaystyle f$ is the finite set $\displaystyle \{y_1,\cdots,y_k\}$. We can put $\displaystyle \epsilon$-balls $\displaystyle B_i$ around each $\displaystyle y_i$ such that $\displaystyle B_1,\cdots,B_k$ are disjoint. Then $\displaystyle E_1,\cdots,E_k$ are measurable sets satisfying (1) and (2), where $\displaystyle E_i=f^{-1}(B_i)$.

Have I misunderstood something here?

7. Re: a question about Lebesgue measurable function

Aren't they all translates of the set of points with rational coordinates?

8. Re: a question about Lebesgue measurable function

Originally Posted by hatsoff
The way I read this it's impossible. Let $\displaystyle f:E\to\mathbb{R}^n$ be a finite-valued Lebesgue measurable function. The codomain of $\displaystyle f$ is the finite set $\displaystyle \{y_1,\cdots,y_k\}$. We can put $\displaystyle \epsilon$-balls $\displaystyle B_i$ around each $\displaystyle y_i$ such that $\displaystyle B_1,\cdots,B_k$ are disjoint. Then $\displaystyle E_1,\cdots,E_k$ are measurable sets satisfying (1) and (2), where $\displaystyle E_i=f^{-1}(B_i)$.

Have I misunderstood something here?
I think that by finite-valued it's meant that the function has range in the reals, as opposed to the extended line.

Aren't they all translates of the set of points with rational coordinates?
Yes, but see Vitali set.

9. Re: a question about Lebesgue measurable function

Originally Posted by Jose27
I think that by finite-valued it's meant that the function has range in the reals, as opposed to the extended line.
Ah, okay. Well how about this: Let $\displaystyle E=\mathbb{Q}$, and define $\displaystyle f:E\to\mathbb{R}$ by $\displaystyle f(x)=0$ for $\displaystyle x=0$ and $\displaystyle f(x)=b$ otherwise, where $\displaystyle |x|=a/b$ in lowest terms.

EDIT: nevermind, that won't work.

10. Re: a question about Lebesgue measurable function

That function has countable range, take $\displaystyle E_b=f^{-1}(b)$. I'm not sure we'll be able to give this function explicitly.

11. Re: a question about Lebesgue measurable function

Originally Posted by Jose27
That function has countable range, take $\displaystyle E_b=f^{-1}(b)$. I'm not sure we'll be able to give this function explicitly.
Why does it matter if the range is countable? All we need is that $\displaystyle f:E\to\mathbb{R}$ be Lebesgue measurable such that for any countable, Lebesgue-measurable partition $\displaystyle \{E_1,E_2,\cdots\}$ of $\displaystyle E$ we have $\displaystyle f$ discontinuous on some $\displaystyle E_k$, right?

12. Re: a question about Lebesgue measurable function

Yes, but if the range is countable, a decomposition similar to the one I gave has the desired properties since the function would be constant in each set.

Your function is discontinous in any open set, but not on any of the measurable ones I gave.

13. Re: a question about Lebesgue measurable function

Lusin's theorem.

14. Re: a question about Lebesgue measurable function

hello,hatsoff,thank you for your reply.I am sorry, Maybe I did not say clearly.The function f what I said is real-valued.

15. Re: a question about Lebesgue measurable function

I've been thinking about this but can't seem to make any progress. Can you prove the statement when instead of a countable family we ask it be finite? (Shame, I can't even prove this, not even for two!).

As an aside, if the sets are asked to be closed then the problem is trivial by Tietze's extension theorem.

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