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Let $f$ be defined as

$$f(x)=\begin{cases} x \sin(\frac{1}{x}) & x\ne 0 \\ 0 & x=0 \end{cases}$$

$f(x)$ is not absolutely continuous so it cannot might not satisfy the Lusin N condition.

Is there a direct proof of that it does not? i.e. I wanted to know how to construct a set of zero measure which does not satisfy Lusin's Condition for this function.

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Myke, I fixed up some formatting and some wording. If that is not your intent, I suppose you will change it :-) –  Aryabhata Sep 15 '10 at 18:59
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@Myke: I believe \sout requires the ulem package, which is not installed. The html way is to use <s>text</s> (or <strike>text</strike>) –  Larry Wang Sep 15 '10 at 20:32
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@Chandru: I put that strikethrough because Myke wanted it. I have rolled back to that change. –  Aryabhata Sep 15 '10 at 20:53
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Qiaochu: Lusin's N Condition means that f maps a set of measure 0 to another set of measure 0 –  Digital Gal Sep 15 '10 at 21:38
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Qiaochu: Correction Lusin's N Condition means that f maps sets of measure 0 to sets of measure 0 –  Digital Gal Sep 15 '10 at 21:55

1 Answer 1

up vote 3 down vote accepted

The function sends each set of measure zero to a set of measure zero.

Let $A\subset\mathbb R$ be a null set (i.e. $A$ has measure 0). Then $f(A)=f(A\cap \{0\})\cup f(A\cap(\mathbb{R}\setminus \{0\}))$. The first set in the union has at most one point, so we need only worry about the second. The set $A\cap(\mathbb{R}\setminus \{0\})$ can be expressed as a countable union of sets of the form $A\cap [a,b]$ with $0\lt a$ or $b\lt 0$, and therefore, since the image of a union is the union of the images, $f(A\cap(\mathbb{R}\setminus \{0\}))$ can be expressed as a countable union of sets of the form $f(A\cap [a,b])$ with $0\lt a$ or $b\lt 0$. For each such $a$ and $b$, $f$ is continuously differentiable in a neighborhood of $[a,b]$, and hence the restriction of $f$ to $[a,b]$ is absolutely continuous (e.g. by the fundamental theorem of calculus for $C^1$ functions). Therefore the image of the null set $A\cap[a,b]$ under $f$ is null. Countable unions of null sets are null, so this shows that $f(A)$ is null.

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+1. But couldn't this argument be simplified just by considering $f$ on sets $[\epsilon, 1]$ for $\epsilon \to 0$? Or am I missing a key point? –  whuber Sep 16 '10 at 2:28
    
To get a countable union and to cover all of R\{0}, you can take for example sets of the form [1/n,n] and [-n,-1/n] ranging over all positive integers n. You can think of the [a,b]s that way, but I preferred to write it without those details. –  Jonas Meyer Sep 16 '10 at 2:32
    
In retrospect, I don't know why I wrote A intersect (R\{0}) rather than A\{0}, but I'm not sufficiently bothered to want to edit. –  Jonas Meyer Sep 16 '10 at 3:06

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