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I want to prove that a function $f:\mathbb{R}^n\rightarrow \overline{\mathbb{R}}$ that is continuous everywhere except for a set $E$ of Lebesgue measure zero is a Lebesgue measurable function. We know that $f$ is not continuous, so there are some open sets ${U_k}$ in $\overline{\mathbb{R}}$ such that $f^{-1}(U_k)$ is not open in $\mathbb{R}^n$. But $f^{-1}(U_k)\subseteq E$, and so $f^{-1}(U_k)$ is still measurable because it has measure zero.
I don't know if this works, because I'm not sure if $E$ contains actually the preimage of those open sets $U_k$.

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    $\begingroup$ Are you using a complete version or non-complete one? Because it sounds wrong if it's not complete, and is easy if it is. $\endgroup$ – Gina Dec 30 '13 at 19:12
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You are using that subsets of measure-zero-sets are again measurable for the Lebesgue measure. This isn't trivial, a measure with this property is called complete and the Lebesgue measure indeed has that property.

Now from the viewpoint of logic, it may be good to start with the fact that $f$ is continuous - well, almost everywhere. So Let $E \subset \Bbb R^n$ be the set of points where $f$ is discontinuous.

Now from the definition, a function is Lebesgue-measurable if the preimage of any set in the Borel-algebra of $ \Bbb R$ is Lebesgue-measurable. As the Borel-algebra on $\Bbb R$ is generated by the open sets, it's enough to see that the preimage of any open set is measurable.

Now show that you can partition the preimage of any open set in an open set and a measurable subset of $E$. From the above argument, this shows that $f$ is Lebesgue-measurable.

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  • $\begingroup$ Depending on your definition of a measurable set, it might be trivial (a set is measurable w.r.t. $\mu$ iff it's measurable w.r.t. its completion). And subsets of measurable sets are not measurable, subsets of measure zero sets are. $\endgroup$ – tomasz Dec 30 '13 at 19:36
  • $\begingroup$ thank you, that's what I wanted to say. I'll edit the specific sentence. $\endgroup$ – benh Dec 30 '13 at 19:38
  • $\begingroup$ Don't forget that the OP is talking about $\mathbb{R}^{n}$, not $\mathbb{R}$. Depend on how you define Lebesgue measure on $\mathbb{R}^{n}$ (just product measure, or its completion) this problem is either false or trivial. $\endgroup$ – Gina Dec 30 '13 at 19:42
  • $\begingroup$ You're right (thank you, added the $n$). But I didn't know that there is an incomplete Lebesgue measure on $\Bbb R^n$. Mine was always complete :) Thanks for pointing that out, so I guess the OP should decide what measure he considers. $\endgroup$ – benh Dec 30 '13 at 19:51
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    $\begingroup$ It's you who have to decide. As Gina pointed out, it depends on your definition of Lebesgue-measure, more precisely on the question whether your Lebesgue-measure can measure subsets of zero-sets. If so, then everything should work. If not, we have a problem, because your statement is false (take any indicator-function of a non-measurable subset of a zero-measure-set). So the question is: is your Lebesgue-measure defined as the product measure of the one-dimensional Lebesgue-measure? $\endgroup$ – benh Dec 30 '13 at 20:38

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