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How do I show that the characteristic function $\chi$ of SVC is not Riemann integrable on $[0,1]$ with no using of measure.

I'd like to show: $U(\chi,P)-L(\chi, P)\geq \frac{1}{2}$

Thanks a lot!

EQJ
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  • Do you have an idea of what $L(\chi, P)$ might be? And where a lower bound for $U(\chi,P)$ might come from? –  Jun 04 '14 at 22:25
  • $L(\chi, P)=0$ for every partition $P$. And for a partition $P_n$ with the border points of $C_n$ we have $U(f,P_n)=\displaystyle\frac{1}{2}+\displaystyle\frac{1}{2^{n+1}}>1/2$. – EQJ Jun 05 '14 at 03:14
  • Sounds good. I don't quite see where $1/2$ comes from, because you probably have a particular instance of SVC in mind (which you did not tell us about). –  Jun 05 '14 at 03:16
  • That follow from the construction of SVC. I see it from here http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set – EQJ Jun 05 '14 at 03:19
  • OK, that's an example that the writer of that article chose to illustrate the set. Generally, saying that something is a SVC (more commonly, "fat Cantor set") carries no information about the measure of the set, other than it's positive. –  Jun 05 '14 at 03:23
  • I know that should be used a concept near to a measure. But I was thinking that there exist a prove similarly to the prove of $\chi_C$ is integrable on $[0,1]$ in which $C$ is the usual Cantor set. In this link http://math.stackexchange.com/questions/18474/riemann-integral-of-characteristic-function-of-cantor-set the first anwers do not use measure. – EQJ Jun 05 '14 at 03:28
  • Indeed I'd like to prove with no measure that $\bar\int_0^1 \chi_C=\displaystyle\frac{1}{2}$. – EQJ Jun 05 '14 at 03:35

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Looks like this has been mostly sorted out in comments. First observation is that the lower sum is always $0$, since the set does not contain any interval. Let's call the set $F$, and the $n$th step of its construction $F_n$.

It is also easy to show that $\inf U\le 1/2$, using $F_n$ to make suitable $P$.

The hard part is to show $\inf U(\chi,P)\ge 1/2$. It's not enough to look at just one $P$. The estimate must be proved for a general $P$.

Take the union of all subintervals $I_j$ from $P$ that intersect the set $F$. We must show that the sum of lengths of these subintervals is at least $1/2$.

To minimize the amount of measure theory involved, you may want to prove that there exists $n$ such that $F_n\subset \bigcup U_j$. This leads to the desired estimate at once, since the $F_n$ is just the union of finitely many intervals whose lengths we know.

A hint for the existence of $n$ as above: let $\delta$ be the smallest length of a partition interval from $P$ that does not intersect $F$. Choose $n$ large enough, all gaps created at the steps $n+1$, $n+2$,... are shorter than $\delta$. Conclude that any interval from $P$ that does not intersect $F$ also does not intersect $F_n$.