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I'm reading a few papers and trying to understand why one paper is supposed to be a strengthening of the other. The bulk of the papers didn't have to do with what follows, but at one point they imply that an a.e. essentially continuous characteristic function is automatically a characteristic function of a set which is a.e. an open set, and this is stronger. I'll make the definitions rigorous:

We say $x \in \mathbb{R}^n$ is a point of $\textit{essential continuity}$ for $f$ (a function on $\mathbb{R}^n$) if for every $\varepsilon > 0$, there exists a neighborhood $U_{x,\varepsilon}$ of $x$ such that $|f(y)-f(x)| < \varepsilon$ for a.e. $y \in U_{x,\varepsilon}$ (hence the "essentially" continuous). I have two main questions (and one side question which is less important):

  1. Suppose the characteristic function $1_E$ is a.e. essentially continuous, that is, for a.e. $x$, $x$ is a point of essential continuity for $1_E$. Then why must $E$ be a.e. equivalent to an open set?

    For this question, I was thinking of letting $\varepsilon < 1$ be fixed, and then the definition of essential continuity implies that for a.e. $x$, there is a neighborhood $U_x$ of $x$ such that for a.e. $y \in U$, we have $|1_E(x)-1_E(x)| < 1$. In particular, this means either both $x$ and $y$ are in $E$, or both are not in $E$. I tried to let my open set $U$ be the union of all the $U_x$ for $x \in E$ which are points of essential continuity, but the problem is there may be points $y$ in each $U_x$ which are not in $E$, and there could be uncountably many $U_x$. Thus the resulting $U$ is not in fact open.

  2. Is there a good example of a set $E$ which is a.e. equivalent to an open set, but $1_E$ is not a.e. essentially continuous?

    I don't really know where to start with this one, but I imagine we require a highly disjoint set, like some sort of fat cantor set. This particular set won't work though, since the fat Cantor set, say $K$, is not equivalent to an open set, but perhaps the complement $[0,1] \setminus K$ will work - I'm stuck on this too though.

  3. I'm curious if we can strengthen 1 and 2: with the hypothesis of 1, why must both $E$ and $E^c$ be a.e. equivalent to open sets? And for 2, does there exist a set $E$ a.e. equivalent to an open set such that $E^c$ is also a.e. equivalent to an open set, and $1_E$ is not a.e. essentially continuous?

    The motivation for 3 is that I believe the first paper's result can be strengthened from a statement about $E$ being a.e. equivalent to an open set, to a statement about both $E$ and $E^c$ being equivalent to an open set. If so, then in order for the second paper to be a true generalization, point 3 must hold.

Any help is appreciated, thanks.

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See this page for how to use Markdown formatting - in particular, you can use *text* to produce italics, instead of resorting to $\textit{text}$. – Zev Chonoles Jun 9 '12 at 19:52
Concerning 3: Since $1_{E^c}=1-1_E$, the essential a.e. continuity of $1_E$ tells you as much about $E^c$ as about $E$. – user31373 Jun 9 '12 at 19:59
That makes sense to me, except your example for 2 won't work for 3. If $E$ is the complement of a fat Cantor set, then $E^c$ is not equivalent to an open set - for 3 I was curious if there was a set $E$ for which both $E$ and $E^c$ are equivalent to open sets, where $1_E$ is not a.e. essentially continuous. Am I missing something? – Ovrf Jun 9 '12 at 21:34
If $E$ is equivalent to an open set $U$ and $E^c$ is equivalent to an open set $V$, then $U$ and $V$ are disjoint sets whose union is $\mathbb R^n$ minus a null set. The characteristic function of $E$ is essentially continuous on $(E\cap U)\cup (E^c\cap V)$, so it is essentially continuous a.e. – user31373 Jun 9 '12 at 21:53
Why is the characteristic function of $E$ essentially continuous on $(E \cap U) \cup (E^c \cap V)$? If this is true, then I'm disappointed - the second paper is not a generalization of the first paper after all, since this would prove we have an if and only if situation. – Ovrf Jun 9 '12 at 22:31
up vote 1 down vote accepted

Re: 1. Taking $\bigcup U_x$ is a good idea, here $x$ should range over essential continuity points that belong to $E$. You have $U_x\subset E$ for all such $x$ (well not really :). Uncountability is not a problem: arbitrary union of open sets is open.

Re: 2. For any open set $E$, the function $1_E$ is not essentially continuous at any point of $\partial E$. Indeed, any neighborhood of a boundary point contains a ball $B\subset E$. So, if the boundary of an open set $E$ has positive measure (e.g., $E$ is the complement of a fat Cantor set), then $1_E$ is not essentially continuous a.e.

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Hi Leonid, thanks for the reply. For 1, I wasn't concerned about openess; of course $U$ as I defined it is open. The issue about uncountability arises because the $y \in U_x$ which do not satisfy the inequality, and hence (when $x \in E$) $y \notin E$. Let $Y_x$ be the set of $y \in U_x$ for which $y \notin E$. Then for an individual $U_x$, $|Y_x| = 0$, but when you union all the $U_X$, the resulting $U$ differs from $E$ by the set $\cup Y_x$, which is an uncountable union and thus might not be measure 0. – Ovrf Jun 9 '12 at 21:31
Ah, your proof for 2 makes sense. The complement of the fat Cantor set was what I had in mind as a possible example, but I only had intuition. Thanks for this. – Ovrf Jun 9 '12 at 21:36
@Ovrf At such exceptional $y\in U_x$ the characteristic function is not essentially continuous. So their set has measure zero after all. – user31373 Jun 9 '12 at 21:47
Yes, I see that now. Thank you Leonid, this proof seems fine to me. All that remains to satisfy my curiosity is my Q3. – Ovrf Jun 9 '12 at 22:29

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