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$f$ is discontinuous on a subset of $\Bbb R$, which one is impossible?

A. empty set
B. rational numbers
C. irrational numbers
D. positive real numbers
E. $\Bbb R$

I have excluded B.C.D.E. If A is the answer then the preimage of null set with a function from $\Bbb R$ to $\Bbb R$ is also a null set. Is it right?

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Is it possible for you to format the question better? Two line breaks make a new paragraph. One line break is ignored, unless two whitespace characters are put at the end of the previous line. –  Asaf Karagila Mar 28 '13 at 10:05

2 Answers 2

up vote 1 down vote accepted

Hint: Recall that the set of points of continuity of a function $\mathbb{R} \to \mathbb{R}$ is a G$_\delta$ set (see here), which means that the set of points of discontinuity is F$_\sigma$. Are any of the sets listed not F$_\sigma$?

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$\Phi$ is both a $F_{\sigma }$ and $G_{\delta }$ set. Isn't it? –  Jebei Mar 28 '13 at 11:23
    
Is it A possible? –  Jebei Mar 28 '13 at 11:27
    
Yes, $\emptyset$ is both an F$_\sigma$ and G$_\delta$ set. (In general, every open subset of $\mathbb{R}$ is both F$_\sigma$ and G$_\delta$, and similarly for closed sets.) Note that $\emptyset$ is the set of discontinuities of any continuous function, such as $f : x \mapsto x$. –  Arthur Fischer Mar 28 '13 at 11:32

All of them are possible.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be $f(x) = \lfloor x \rfloor$.

$f$ is discontinuous.

$\mathbb{Z}$ is a null subset of $\mathbb{R}$ since it is countable. $f^{-1}(Z) = \mathbb{R}$. So $E$ is possible.

The positive $\mathbb{Z}^+$ is null since it is countable. $f^{-1}(\mathbb{Z}^+) = \mathbb{R}^+$. $D$ is possible.

$\{\frac{1}{2}\}$ is null. $f^{-1}(\frac{1}{2}) = \emptyset$. So $A$ is possible.

Define $g(x) = \begin{cases} 1 & \quad x \in \mathbb{Q} \\ 0 & \quad x \notin \mathbb{Q} \end{cases}$

$g$ is not continuous. $\{1\}$ is null, $g^{-1}(1) = \mathbb{Q}$. So $B$ is possible.

Similarly, $\{0\}$ is null. $g^{-1}(0) = \mathbb{R} - \mathbb{Q}$. So $C$ is possible.

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I should mention that I assumed null meant measure 0. But this would work for any notion of "null" in which countable sets are "null". –  William Mar 28 '13 at 10:17
    
Null set here means empty set. Not a 0 out measure set. Is it possible? –  Jebei Mar 28 '13 at 11:07
    
@frame99 The preimage of the empty set under any function is the empty set. –  William Mar 28 '13 at 11:10

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