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I came across the following problem in my self-study:

Let $f:[0,1] \rightarrow [0,1]$ be the Cantor (ternary) function, and let $g(x) = f(x) + x$. Then:

(a) $g$ is bijection from $[0,1]$ to $[0,2]$, and $h = g^{-1}$ is continuous.

(b) If $C$ is the Cantor (ternary) set, $m(g(C))=1$.

(c) It follows that $g(C)$ contains a Lebesgue nonmeasurable set $A$. Let $B = g^{-1}(A)$, and show that $B$ is Lebesgue measurable but not Borel.

(d) There exist a Lebesgue measurable function $F$ and a continuous function $G$ on $\mathbb{R}$ such that $F \circ G$ is not Lebesgue measurable.

I think I can prove (a), but I am not having any nice ideas on how to proceed for the remaining (3) parts, and I am interested to see if anyone visiting knows how to tackle this interesting exercise.

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For (b) consider $[0,2]-g([0,1]-C)$ which is a countable disjoint union. –  Tim kinsella Feb 1 '12 at 6:58
    
For (c) note that $B$ is a subset of a null set. Use the continuity of $h$ to show that $B$ can not be Borel. –  Tim kinsella Feb 1 '12 at 7:07
    
sorry, i meant for (b) note that $g(C)= [0,2]-g([0,1]-C)$ and that $g([0,1]-C)$ is a countable disjoint union of intervals. –  Tim kinsella Feb 1 '12 at 7:23

1 Answer 1

up vote 2 down vote accepted

For (b), the key observation is that, as $g$ is continuous, the image of an interval is an interval, and so $m(g((a,b)))=g(b)-g(a)$.

Then, as Tim said, we can use that the complement of $C$ is a countable disjoint union of open intervals $C_k$ whose measures add to 1, and that on these intervals we have $f=0$. So, if $C_k=(a_k,b_k)$, then $$ m(g(C_k))=g(b_k)-g(a_k)=f(b_k)-f(a_k)+b_k-a_k=b_k-a_k=m(C_k). $$ So \begin{eqnarray} 2-m(g(C))&=m([0,2]\setminus g(C))=m(g(\bigcup_k C_k))=m(\bigcup_k g(C_k))=\sum_k m(g(C_k)) =\sum_k m(C_k)=1, \end{eqnarray} i.e. $m(g(C))=1$.

For (c), it is an established result that any set of positive measure contains a non-measurable subset. So let $A\subset g(C)$ be non-measurable, and put $B=g^{-1}(A)$. Since $B\subset C$ and $C$ is a null-set, $B$ is also a null-set, and it particular it is measurable. But it is not Borel: since $g^{-1}$ is continuous, the pre-image of a Borel set is Borel; as $A=g(B)$, this would make $A$ Borel and thus measurable, which it's not.

Finally, for (d), the key is that $$ 1_A=1_{g(B)}=1_B\circ g^{-1}. $$ The characteristic $1_A$ is not measurable since $A$ isn't. But $B$ is measurable so $1_B$ is, and $g^{-1}$ is continuous. So the composition of measurable functions may fail to be measurable, even if one of them is continuous.

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