12
$\begingroup$

Say $E \subset [0,1]$ is a null set. Let $f: [0,1] \rightarrow [0,1] $. Do you think $f(E)$ is a null set or not? Just being curious.

(DEF): A set $A$ is null if given any $\epsilon > 0$, there exists a sequence of intervals $\{I_n\}_{n\geq1}$ such that

$$ A \subseteq \bigcup _{n=1}^{\infty}I_n$$ and $$ \sum |I_n| < \epsilon $$

if $f$ is continuous, is $f(E)$ nullset or not?

$\endgroup$
4
  • 1
    $\begingroup$ What is a null set? $\endgroup$
    – Matemáticos Chibchas
    Aug 30, 2013 at 20:05
  • 1
    $\begingroup$ Depends on $f$. If $f$ is Lipschitz (absolutely continuous should suffice), then $f(E)$ is a null set. In general, it need not be one. $\endgroup$
    – Daniel Fischer
    Aug 30, 2013 at 20:06
  • $\begingroup$ With null set you mean empty, right? Because Wikipedia thinks differently (from what I thought). $\endgroup$
    – JMCF125
    Aug 30, 2013 at 20:11
  • $\begingroup$ In the future, please use more descriptive titles. I've edited this one. $\endgroup$
    – Nate Eldredge
    Sep 2, 2013 at 2:54

2 Answers 2

Reset to default
15
$\begingroup$

If we do not put any conditions on $f$, the answer is "not necessarily."

For example the Cantor set is a null set, but there is a one-to-one onto mapping $f$ of the Cantor set to the unit interval.

$\endgroup$
4
  • 5
    $\begingroup$ And there is a (uniformly) continuous, at most $2$-$1$ mapping $f$ of the Cantor set onto $[0,1]$ as well, so even uniform continuity isn’t enough. $\endgroup$
    – Brian M. Scott
    Aug 30, 2013 at 20:17
  • 1
    $\begingroup$ Note this answer doesn't exactly address the question, which is about continuous mappings. A one-to-one onto mapping of the Cantor set onto the unit interval is necessarily discontinuous (as the Cantor set and the unit interval are compact Hausdorff, a continuous bijection would be a homeomorphism, which is absurd.) $\endgroup$
    – Nate Eldredge
    Sep 4, 2013 at 0:12
  • $\begingroup$ This answers the original version of the question, which didn't mention continuity even; supplemented now with Brian and @Nate's comments, more bases are covered. $\endgroup$
    – Jonas Meyer
    Sep 4, 2013 at 0:14
  • 2
    $\begingroup$ The question changed, perhaps because of my answer. I was not notified. $\endgroup$
    – André Nicolas
    Sep 4, 2013 at 0:18
15
$\begingroup$

To have that property it is not enough to be just continuous as many above have pointed out. The Devil's ski slope homeomorphism from $[0,1]$ to $[0,2]$ maps the standard cantor set to a fat cantor set (of measure $1$). We can map $[0,2]$ back to $[0,1]$ by composing with the linear transformation $\frac{x}{2}$ (which maps the measure $1$ cantor set in $[0,2]$ to a set of measure $\frac{1}{2}$ in $[0,1]$). So even being a homeomorphism is not enough. What you seek is for your function to have the Luzin N property (http://en.wikipedia.org/wiki/Luzin_N_property). Absolutely continuous functions, Lipschitz functions, and $C^{1}$ diffeomorphisms are types of functions with this property.

$\endgroup$
1
  • 3
    $\begingroup$ Incidentally, $C^1$ functions on $[0,1]$ are Lipschitz, and Lipschitz functions on $[0,1]$ are absolutely continuous. To see that absolute continuity is strictly stronger than Luzin N, note that if a continuous function on $[0,1]$ is absolutely continuous on each interval $[\varepsilon,1]$ with $\varepsilon>0$, then it has the Luzin N property on $[0,1]$, but need not be absolutely continuous. $\endgroup$
    – Jonas Meyer
    Sep 4, 2013 at 0:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.