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Recall Lusin's Theorem:

Let $f$ be a real-valued measurable function on $E=[0,1]$. Then, for each $\epsilon > 0$, there is a continuous function $g$ on $\mathbb{R}$ and a closed set $F$ contained in $E$ for which $f=g$ on $F$ and $m(E\backslash F) < \epsilon$.

I was wondering if $f$ itself ever has to be continuous at any point (as a function on $E$)? I figured that it would at least need to be continuous on the interior of $F$, where it actually equals $g$ (I realize it technically equals $g$ on the boundary of $F$ as well, but we don't know what's going on outside $F$, so we can't necessarily say it's continuous on the boundary, can we?), because there it's coinciding with a function that is known to be continuous.

Is there any more to this, and if so (and even if not), what would a proof look like?

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    $\begingroup$ What about $f=1_\Bbb {Q} $? $\endgroup$
    – PhoemueX
    Nov 4, 2015 at 20:41
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    $\begingroup$ $f$ needs to be continuous on the interior of $F$, right. But the interior of $F$ can be empty. $\endgroup$ Nov 4, 2015 at 20:42
  • $\begingroup$ There is another way to think about Lusin's theorem: $f$ is itself continuous on the closed set $F$ (i.e, relative to that set) but is not necessarily continuous at any point of $F$ since that kind of continuity pays attention to points outside $F$ which we do not control (other than stuffing them into a set of small measure). $\endgroup$ Nov 4, 2015 at 21:30
  • $\begingroup$ @DanielFischer, I just realized I forgot an important stipulation that might make what you suggested not feasible. See the edit I made above. $\endgroup$
    – user100463
    Nov 5, 2015 at 2:37
  • $\begingroup$ @PhoemueX, I realized I had forgotten to define $E=[0,1]$. Can you still use $1_{\mathbb{Q}}$ there and get $m(E\backslash F)<\epsilon$? $\endgroup$
    – user100463
    Nov 5, 2015 at 2:40

1 Answer 1

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Since continuity (at a point) is a local property - whether $f$ is continuous at $x_0$ depends only on the values of $f$ on arbitrarily small neighbourhoods of $x_0$ - it indeed follows that $f$ must be continuous on the interior of $F$, if $f$ coincides with the continuous function $g$ on $F$.

But of course $F$ can have empty interior, so that doesn't imply that a measurable function needs to be continuous at any point.

And indeed, there are measurable functions that are nowhere continuous.

The characteristic function of $\mathbb{Q}\cap [0,1]$ is of course measurable and nowhere continuous.

However, by changing the values on a null set, we obtain the everywhere continuous function $0$. And conversely, every measurable function can be changed to a nowhere continuous measurable function by changing the values on a null set.

More interesting is the fact that there are measurable functions that are nowhere continuous and cannot be made continuous anywhere by changing the values on a null set.

The "simplest" example of such a function is the characteristic function of a measurable set $A$ with the property that $0 < m(A\cap I) < m(I)$ for all non-degenerate intervals $I$. Then every function $f$ that is a.e. equal to $\chi_A$ has points $x,y$ with $f(x) = 0$ and $f(y) = 1$ in every non-degenerate interval, hence cannot be continuous anywhere.

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    $\begingroup$ but my problem is that if $F$ has empty interior, then $m(F) = 0$, right? So, then we can't have $m(E\F)=0$ because then we wouldn't be satisfying Lusin's Theorem. What I am asking is whether it is possible for a function to satisfy Lusin's Theorem but not be continuous anywhere, and if not, I'd like to see a proof. $\endgroup$
    – user100463
    Nov 5, 2015 at 17:15
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    $\begingroup$ No, $F$ can have positive measure even when it has empty interior. We can't have $m(E\setminus F) = 0$ then, but that would be more than Lusin's theorem asserts. It only says that for all $\epsilon > 0$ we can find an $F$ with $m(E\setminus F) < \epsilon$. To construct closed $F$ with empty interior such that $m(E\setminus F) < \epsilon$, you can for example take an enumeration $(r_n)$ of the rationals in $E$, and let $U = E\cap \bigcup (r_n - \epsilon\cdot 2^{-n-3}, r_n + \epsilon\cdot 2^{-n-3})$. Then $U$ is an open set containing $E\cap \mathbb{Q}$, so $F = E\setminus U$ is a closed set … $\endgroup$ Nov 5, 2015 at 17:25
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    $\begingroup$ with empty interior, and since $$m(U) \leqslant \sum_{n = 0}^{\infty} m((r_n - \epsilon\cdot 2^{-n-3}, r_n + \epsilon\cdot 2^{-n-3})) = \sum_{n = 0}^{\infty}\epsilon\cdot 2^{-n-2} = \frac{\epsilon}{2},$$ it follows that $m(E\setminus F) = m(U) < \epsilon$. For another (not entirely dissimilar) construction of such sets, you could look up "fat Cantor sets". $\endgroup$ Nov 5, 2015 at 17:25
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    $\begingroup$ What do you mean, where would $f$ have to be continuous? We can have measurable but nowhere continuous $f$ (and such that the discontinuity couldn't be eliminated by changing $f$ on a null set). Lusin's theorem says that there are continuous functions that coincide with $f$ on sets of measure arbitrarily close to that of $E$. But that doesn't mean $f$ needs to be continuous anywhere. $\endgroup$ Nov 5, 2015 at 17:33
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    $\begingroup$ $\mathbb{Q}\cap E$ is dense in $E$. That means every nonempty open set in $E$ contains elements of $\mathbb{Q}\cap E$. But in our construction, we have $\mathbb{Q}\cap E \subset U$, hence $F = E\setminus U$ contains no rational, and therefore it cannot contain any nonempty open set. $\endgroup$ Nov 5, 2015 at 17:43

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