In short: If $f$ is continuous on a measure one set, is there a function $g=f$ a.e. such that a.e. point is a point of continuity of $g$?

Now more carefully, with some notation: Suppose $(X, d_X)$ and $(Y, d_Y)$ are metric spaces, with Borel sets $\mathcal B_X$ and $\mathcal B_Y$, respectively. Let $f:(X, \mathcal B_X) \rightarrow (Y, \mathcal B_Y)$ be a measurable function and let $\mu$ be a probability measure on $(X, \mathcal B_X)$.

Say that $f$ is almost continuous if there is a $\mu$-measure-one Borel set $D \subseteq X$ such that $f$ is continuous on $D$, i.e., the restriction $f|_D:D \rightarrow Y$ is a continuous function where $D$ is given the subspace topology.

As usual, we say that $x \in X$ is a point of continuity (for $f$) if, for every Cauchy sequence $(x_n)_{n\in\mathbb{N}}$ that converges to $x$, we have that $(f(x_n))_{n \in \mathbb{N}}$ is a Cauchy sequence that converges to $f(x)$.

Let $C$ be the set of all points of continuity of $f$. (A classic result [Kechris, I.3.B Prop 3.6] shows that $C$ is a $G_\delta$ set.) Say that $f$ is almost everywhere (a.e.) continuous if $C$ is a $\mu$ measure-one set., i.e. $\mu$-a.e. point is a point of continuity.

Finally, say that a measurable function $g$ is a version of $f$ if $f=g$ $\mu$-a.e.

Clearly, if $f$ is a.e. continuous, then it is almost continuous on the set $C$. The converse does not hold in general: Consider for $f$ the indicator function for the rationals in $[0,1]$. Then $f=0$ on the irrationals, a Lebesgue-measure-one set, but $f$ is discontinuous everywhere.

However, $g=0$ is a version of $f$ and $g$ is a.e. continuous. Which raises the question:

If $f$ is almost continuous, is there a version $g$ of $f$ such that $g$ is a.e. continuous?

If it is helpful, one may assume that both $X$ and $Y$ are also complete and separable, i.e., Polish.

Note that by a result of Kuratowski [Kechris, I.3.B Thm. 3.8], if $Y$ is complete, we can, possibly changing versions, assume, without loss of generality, that $f$ is continuous on a $\mu$-measure-one $G_\delta$-set $D$. (In the example above, $f$ is almost continuous on the irrationals, which are of course $G_\delta$.)

[Kechris: "Classical Descriptive Set Theory" 1995]

  • $\begingroup$ I think the answer is yes. For each point not in the measure one set, let its value be the lim sup of all the values in a ball around it (restricted to the measure one set) as the radius goes to zero. Then by the definition of continuity this should work. I'll try to think about the details and write an answer later if no one else does. $\endgroup$
    – Jason Rute
    Oct 26 '13 at 19:28
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    $\begingroup$ Thanks. However we are not assuming that Y is the real numbers and so there is not necessarily a natural ordering. Also, even when Y is the real numbers I believe that for many points not in the measure one set the limit will not exists (i.e. will be $\infty$) $\endgroup$ Oct 26 '13 at 21:55
  • $\begingroup$ I see my solution/comment above was wrong even if $Y=[0,1]$ (even when the lim sup is not $\infty$). I still think the answer to your question is yes (when $X$ and $Y$ are Polish), but whatever the proof is, it is going to be technical I think. You should add that $X$ and $Y$ are Polish spaces (I assume this is what you meant---since you cited Kechris). I'll add a "descriptive set theory" tag---since in my mind descriptive set theory is anything technical involving Polish spaces. :) $\endgroup$
    – Jason Rute
    Oct 27 '13 at 1:44

As Jason and Gerald predicted, the answer is yes for Polish $X, Y$. (Indeed, it is sufficient for $X$ to be merely separable and metrizable and for $Y$ to be merely complete and metrizable.)

As Nate observed, we may assume that $D$ is $G_\delta$. Here is the key fact:

Lemma. If $D \subseteq X$ is a nonempty $G_\delta$-set then there is a Borel map $h:X \to D$ such that $\lim_{x\to x_0} h(x) = x_0$ for every $x_0 \in D$.

Suppose $D = \bigcap_{n\lt\omega} U_n$, where $(U_n)_{n\lt\omega}$ is a descending sequence of open sets such that $U_n \subseteq \bigcup_{x_0 \in D} B(x_0,1/(n+1))$. Any Borel retraction $h:X \to D$ with the property that if $x \in U_n \setminus U_{n+1}$ then $d(h(x),x) \lt 1/(n+1)$ will be as required. By definition, it is always possible to find a suitable $h(x) \in D$ for each $x \in U_0 \setminus D$. To ensure that $h$ is Borel, fix an enumeration $(d_i)_{i \lt \omega}$ of a countable dense subset of $D$ and, if $x \in U_0 \setminus D$, define $h(x)$ to be the first element in this list that matches all the necessary requirements. (We must have $h(x) = x$ for $x \in D$ and it doesn't matter how $h(x)$ is defined when $x \notin U_0$ so long as the end result is Borel.)

Now, if $f:X \to Y$ is Borel and continuous on $D$, then $g = f\circ h$ is a Borel function that agrees with $f$ on $D$ and $$\lim_{x \to x_0} g(x) = f(\lim_{x \to x_0} h(x)) = f(x_0) = g(x_0)$$ for all $x_0 \in D$.

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    $\begingroup$ Actually, Polish is too strong: this works if $Y$ is complete (but not necessarily separable) and $X$ is separable (but not necessarily complete). Gerald's first answer generalizes to show that completeness of $Y$ is necessary. I made much use of the separability of $X$ but I don't know to what extent it is necessary. $\endgroup$ Oct 27 '13 at 17:35
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    $\begingroup$ So ... you first discard all $\mu$-null open sets. To get the answer required, at the end we have to extend $g$ to those discarded sets as well, still retaining continuity at every point of $D$. $\endgroup$ Oct 27 '13 at 19:27
  • $\begingroup$ @Gerald: Since $X$ is second-countable this is not a problem. I'm replacing $X$ by a closed subset of $X$ of $\mu$-measure one wherein all nonempty open sets are $\mu$-postive. $\endgroup$ Oct 27 '13 at 19:49
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    $\begingroup$ But there is still a small thing to do, to get $g$ defined on the original space (the one we have before the open set is deleted). $\endgroup$ Oct 27 '13 at 19:52
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    $\begingroup$ @Gerald: Fixed. $\endgroup$ Oct 27 '13 at 20:31

first answer.
As stated ($X, Y$ merely metric spaces), NO.
(Remark: we may as well take $D=Y$ and $f$ the identity on $D$: if we can do that case, then we can apply it to get the general case.)

Let $X=\mathbb R$, let $Y=\mathbb Q$, define $\mu$ a probability measure with positive measure at each rational point of $X$, the rest of $X$ measure zero. Define $f : \mathbb R \to \mathbb Q$ by $f(x)=x$ for rational $x$ and $f(x)=0$ otherwise. Then $f$ restricted to $\mathbb Q$ is continuous, and that is a set of measure one. Suppose $g : \mathbb R \to \mathbb Q$ is an extension of $f$ and every point of $\mathbb Q$ is a point of continuity of $g$. (A set of measure one must contain all rationals.) As noted in the question (for $\mathbb R$ or even for Polish space, it is presumably older than Kechris), the set $E$ of points where $g$ is continuous is a $G_\delta$. In particular, since $\mathbb Q$ is not a $G_\delta$ in $\mathbb R$, we see that $E$ contains at least one irrational $u$. But if $g$ is continuous at $u$ and $g(x)=x$ for all rational $x$, then $g(u)=u$, oops.

Had the second space $Y$ been the completion $\mathbb R$ of $\mathbb Q$, then $g$ would indeed have been a version and a.e. continuous (even continuous everywhere).


If $X$ and $Y$ are Polish and $Y$ is compact, then YES. (I think my proof can be fixed to handle the noncompact setting, but I don't see how right now.)

My proof involves this lemma.

Lemma. Assume that $f(x)$ is $0,1$-valued and continuous when restricted to a measure one set $D$. Then there are two disjoint open sets $U$ and $V$ such that $\mu(U \cup V)=1$ and $f = 1_U$ $\mu$-a.e.

Proof of Lemma. By continuity on $D$, there are open sets $U',V'$ such that $f(U' \cap D)=1$ and $f(V' \cap D)=0$. Now, since we are in a separable space, let $U' = \bigcup_{i=0}^\infty B_i$ where $B_i$ is a ball such that $\overline {B_i} \subseteq U'$. Similarly, let $V'= \bigcup_{i=0}^\infty C_i$ where $C_i$ is a ball such that $\overline {C_i} \subseteq U'$. Now, we will construct $U \subseteq U'$ and $V \subseteq V'$. Let $U = \bigcup_i B_i \smallsetminus (\overline{C_0} \cup \ldots \cup \overline{C_i})$ and $V = \bigcup_i C_i \smallsetminus (\overline{B_0} \cup \ldots \cup \overline{B_i})$.

Now, consider $x\in D$. If $f(x)=1$, then $x\in B_i$ for some $i$ and $x\notin \overline{C_0} \cup \ldots \cup \overline{C_i} \subseteq V'$. Hence $x\in U$. Similarly, if $f(x)=0$, then $x\in V$. This completes the lemma. $\square$

Proof for $Y=\{0,1\}^\mathbb {N}$.

Now I will prove the main result when $Y$ is the Cantor space $\{0,1\}^\mathbb{N}$. Using the lemma, let $U_n$ and $V_n$ be the open sets corresponding to the $n$th bit of $f(x)$. Then let the $n$th bit of $g(x)$ be $1$ if $x\in U_n$, and $0$ otherwise. Every $x$ in $D$ is a point of continuity of $g$ and $f=g$ a.e.

Proof for compact Polish $Y$.

The idea is to treat $Y$ like Cantor space. Let $\mu_f$ be the pushforward measure. Let $\{B_n\}$ be a countable basis of balls in $Y$ with $\mu_f$-null boundary. We may assume there are no $x \in D$ such that $f(x)$ is on one of these null boundaries. Then by the previous part, there is some $h\colon X \rightarrow \{0,1\}^\mathbb {N}$ such that every $x$ in $D$ is a point of continuity and such that for all $x\in D$ the $n$th bit of $h(x)$ is 1 iff $f(x) \in B_n$.

It just remains to construct $g$ from $h$. We can do so with a continuous algorithm. If the $n$th bit of $h(x)$ is $1$ then make $g(x) \in \overline{B_n}$ if it is $0$ then make $g(x) \in Y \smallsetminus \overline{B_n}$. Since $Y$ is compact, this continuous algorithm either converges to a point or it fails at a finite stage (by fail, I mean there is no such $g(x)$ satisfying the requirements so far). If it fails, then just pick a value consistent with the last consistent stage (if one picks this value effectively then this ensures the map from $h(x)$ to $g(x)$ is continuous).

Remark 1. Note that "almost continuous" can mean different things to different mathematicians. See this definition of almost continuous.

Remark 2. Many of the ideas in this answer come from my paper Computable randomness and betting for computable probability spaces which I think might interest the OP (since I know they are interested in the computable setting), although I am sure the ideas are standard in the non computable setting.


second answer:
Suppose we add some good conditions for $X$ and $Y$. Complete or Polish or something. Then it would suffice to prove this:

LEMMA. Let $X, Y$ satisfy (conditions to be determined). Let $D \subseteq X$ have the subspace topology. Suppose $f : D \to Y$ is continuous. Then there exists $g : X \to Y$ such that: (a) $g(x)=f(x)$ for all $x \in D$; (b) $g$ is continuous at each point of $D$.

That would be a result of point-set topology. And it would imply the result we want. Presumably topologists already know this result, together with the conditions to impose. For example, it should be easy to do the case $X=Y=[0,1]$.


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