Vtyjkyuk

Recently, out of curiosity, I looked up the list of questions for Princeton generals, and one caught my attention:

Can you construct a measurable set on the interval $[0; 1]$ such that its intersection
with any subinterval of $[0; 1]$ has measure neither $0$ nor equal to the measure of
the whole subinterval? If so, is the indicator (characteristic) function of that set Riemann
integrable.

Second part of the question is relatively easy to answer, but first I was not able to figure out.

So the final question is: how can one construct such a set?

Yes, such a set exists. Walter Rudin has constructed one in the following paper:

Rudin, W. "Well-distributed measurable sets." The American Mathematical Monthly 90.1 (1983): 41-42.

In his own words:

Theorem. There is a measurable set $Asubseteq I=[0,1]$ such that
$$0<m(Acap V)<m(V)$$
for every nonempty open set $Vsubseteq I$.

Proof. Let CTDP mean: Compact Totally Disconnected subset of $I$, having Positive measure. Let $I_n$ be an enumeration of all segments in $I$ whose endpoints are rational. Construct sequences $A_n$, $B_n$ of CTDP's as follows:

Start with disjoint CTDP's $A_1$ and $B_1$ in $I_1$.

Once $A_1,B_1,ldots,A_n-1,B_n-1$ are chosen, their union $C_n$ is CTD, hence $I_nsetminus C_n$ contains a nonempty segment $J$, and $J$ contains a pair $A_n,B_n$ of disjoint CTDP's. Continue in this way, and put
$$A=bigcup_n=1^infty A_n.$$
If $Vsubseteq I$ is open and nonempty, then $I_nsubseteq V$ for some $n$, hence $A_nsubseteq V$ and $B_nsubseteq V$. Thus
$$0<m(A_n)leq m(Acap V)<m(Acap V)+m(B_n)leq m(V);$$
the last inequality holds because $A$ and $B_n$ are disjoint. Done.