Could anybody give a hint me how to construct a measurable function $f:[0,1]\rightarrow [0,1]$ such that:
$$\forall_{0\leq a < b \leq 1}: f(a,b)=[0,1]$$
I tried to define a sequence of linear functions $g_n$ that would "vertically cover" more and more of $[0,1]$ like so:
$g_0(x)=x$
$g_1(x) = \begin{cases}2x \text{, for } x\in[0,\frac{1}{2}]\\ 2x-1 \text{, for } x\in[\frac{1}{2},1]\end{cases}$
$...$
$g_n(x) = \begin{cases}2^nx \text{, for } x\in[0,\frac{1}{2^n}]\\ 2^nx-1 \text{, for } x\in[\frac{1}{2^n},\frac{2}{2^n}]\\ ...\\ 2^nx - (2^n-1)\text{, for } x\in[\frac{2^n-1}{2^n},1]\end{cases}$
Then I hoped for such a sequence, of clearly measurable functions, to converge pointwise, so that I could define $f(x)=\lim_{n\rightarrow\infty}g_n(x)$. It seems however, that $g_n$ will not converge and I am stuck looking for a different sequence. Or perhaps my approach is flawed from the very beginning? I will be gratefull for any inights or suggestions.
from Hot Weekly Questions - Mathematics Stack Exchange
Sioux
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