does there exist a monotone increasing function a(alpha):[0,1] to R such that

integrating f on d(a) over [0,1] = sum of ( f(1/n)/(2 to the power n))

for every f belonging to R(a;0,1)?

If such an a exists, it must equal

a(\alpha)=sum_{n\ge \frac{1}{\alpha}} 2^{-n}.

Indeed, for all continous f\in\C[0,1],
the definition of the Riemann-Stieltjes-Integral gives
for the partial sum for a partition \{ x_i \} with
intermidiate values \zeta_i\in(x_i,x_{i+1}) 

\int d(a) f = 
\sum_i f(\zeta_i) (a(x_{i+1})-a(x_i)) =
\sum_{n\in\I} \frac{f(\zeta_i)}{2^n}

with some index set \I \subset \N.

If \sigma=max x_{i+1}-x_i is the mesh size,
then \I will contain at least all naturals <1/\sqrt{\sigma} (plus some more).

The rest of the proof is standard.

Since the continous functions are dense within the Riemann-integrable ones,
we have pinned-down a.

It has the desired property for all f>=0 (semi-continuity arguments).

Decompose f as f_{+}-f_{-} with f_{\pm}>=0 proves the property for all f,
for which both \sum f_{\pm}(1/n)/2^n converge (absolute convergence).

Remains the question whether there exist a-integrable f,
for which \sum f(1/n)/2^n does converge, but not absolutely.

I am sure, if \sum f(1/n)/2^n does converge, but not absolutely,
that you can define a clever partition, such that the
partial sum diverges.


If you find this answer appropriate,
please donate your $12 to charity.

Bye Ben