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I recently did an exercise to show that a monotone function $f:X→ℝ $ is Borel measurable (it even only asked for Lebesgue measurability).

On the other hand, the inverse of the Devil's Staircase function created from the ternary Cantor set is not Borel measurable (stated but not proven in class so not sure about this); does this mean that this function(after choosing the image of the Cantor set appropriately) is not monotone? Is it not the limit of monotone functions and therefore (weakly) monotone? $\newcommand{\cantor}{\mathcal{C}}$

Edit: By 'inverse of the Devil's Staircase', function $g$ defined as follows(I am largely copying from my handwritten lecture notes).

First note that for $x$ written in ternary, if $x∈ \cantor$ then it has no 1s in its ternary expansion, so it can be written $x = \sum_1^∞ \frac{ε_n}{3^n}$ for $ε_n∈ \{ 0,2\}$ (which is effectively a unique representation); then the Devil's Staircase $f$ maps this to

$$ f(x) = \sum_1^∞ \frac{ε_n/2}{2^n}$$

To obtain the 'inverse', first write some $y$ in the image as $\sum_1^∞ \frac{ε_n}{2^n}$ for $ ε_n ∈ \{0,1\} $. If $y$ is not of the form $\frac{c}{2^n}$ for some naturals $c,n$, define

$$g(x) = \sum_1^∞ \frac{2ε_n}{3^n}$$

This gives $gf(x) = x$ for $x$ in $\cantor$ and $fg(y) = y$ for $y$ not of the form $\frac{c}{2^n}$. For $y$ of that form, set $g(y) = \sup\{ g(y_0) : y_0 < y \}$ to try and make it monotone.

Now $g$ was claimed to be non-(Borel) measurable, which I can believe for now. But it seems to me that it is monotone, which can't be by the above exercise. Where does it fail to be monotone?

Calvin Khor
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    The Devil's Staircase function is very non-injective. How do you get an inverse for it? – hmakholm left over Monica Apr 15 '14 at 12:07
  • Sorry, I don't mean that I have an inverse in the strict sense but I will edit my post to make this clearer. – Calvin Khor Apr 15 '14 at 12:13
  • Interesting! (Somewhat quick glance-no time) I'm thinking to argue $g$ may be monotone when we restrict $g$ to the set of (positive) dyadic rationals - I think this follows since for $y_{1}=\frac{c}{2^{m}}\leq\frac{d}{2^{n}}=y_{2}$ (looks like $n=m$ w.l.o.g. is ok) implies ${g(y_{0}):y_{0}<y_{1}}\subset{g(y_{0}):y_{0}<y_{2}}$ yielding $g(y_{1})\leq g(y_{2})$. Lastly, overall monotonicity should fail, I'm thinking, because the multiple of $2$, in $g(y)$, disrupts the ternary expansion - i.e, for $y$, $g(y)\geq g(y')$, where $y'$ may, or may not, be a dyadic rational, need $not$ be true. – Procore Oct 14 '16 at 19:18
  • ...same for increasing/decreasing case. In other words, for any $y$ in the image of $f$, $g(y)$ may be less than (or equal to) $g(y')$, for some dyadic rational $y'$, but $g(y)$ may be greater than (or equal to) $g(y'')$, for some dyadic rational $y''$ with $y'\neq y''$ as one of several cases. Maybe I'm completely wrong, however, at a somewhat quick glance, I'm thinking this might prove what you want? – Procore Oct 14 '16 at 19:22
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    Maybe you mean $g(x)=f(x)+x$ where $f(x)$ is the devil staircase. Then g is continuous injective but it's inverse is not measurable – miraunpajaro Oct 21 '19 at 12:43

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