Riemann-Stieltjes Integral Define:
Partition $P = \{x_0,\dots,x_n\}$
$L(P) = \sum_{i=1}^n (\alpha(x_i)-\alpha(x_{i-1}) \inf (f(x): x \in [x_{i-1}, x_i])$
$U(P) = \sum_{i=1}^n (\alpha(x_i)-\alpha(x_{i-1}) \sup (f(x): x \in [x_{i-1}, x_i])$
$L(P) \le I \le U(P)$. I is the Riemann-Stieltjes Integral.
Using this definition, and assume f(x) is a continues function then:
1)  $\int_0^1 f(x)2^{-[x]} d(2^{[x]}) = 0.5f(1)$
2)  $\int_1^2 f(x) d(x[x+3]) = 4 \int_1^2 f(x) dx + 2f(2)$
My attempt  to 1)
$\sum_{i=1}^n  2^{[x_i]} – 2^{[x_{i-1}]} = 2^1 – 2^0 = 1$
$L(P) = 1 \cdot \inf (f(x)2^{[-x]}) = 1 \cdot 2^0f(0) = f(0)$
$U(P) = 1 \cdot \sup (f(x)2^{[-x]}) = 1 * 2^{-1}f(1) = 0.5f(1)$
Clearly my L(P) is wrong but I’m not sure how/why? And I’m completely stuck on the second question. Can someone please help me? I may be using the definition incorrect. 
NB: I know nothing about the extreme value theorem etc. Basically no other knowledge apart from this definition! Lol 
Question 2: If $f’(x)$ is continuous show
$2 \int_a^b f(x)d(f(x)[x]) = [f^2(x)[x]] (a,b) + \int_a^b f^2(x)d[x]$
My attempted solution:
Using integration by parts:
2(fx)f(x)[x] between (a,b) – int (a,b) f(x)[x]df(x)
= 2(fx)f(x)[x] between (a,b) – int (a,b) f(x)[x]f’(x)dx
Not sure how to get that second term!
Question 3: How do you start to calculate these:
$\int_1^n x^3 d([x]/x^2)$
$\int_1^\infty  d([\log x]/x^2)$
$\int_1^n x(x-1) d(1/[x])$
 A: I suppose that by $[x]$ you mean the floor function.
$\Delta(P)$ denotes the norm of partition, i.e., the maximal length of an interval of the partition $P$. 
1) Using the fact that $2^{-[x]}$ is either 1 (for $x<1$) or $1/2$ (for $x=1$) you can see that for any partition such that $\Delta(P)\le\varepsilon$ you get $U(P)\le \frac12 \sup\{f(x); x\in [1-\varepsilon,1]\}$ and $L(P)\ge \frac12 \inf\{f(x); x\in (1-\varepsilon,1)\}$. Both quantities tend to  $\frac{f(1)}2$ as $\varepsilon\to 0^+$. (Using continuity of $f$.)
2) By $x[x+3]=4x+x[x-1]$ (which holds for $x\in[1,2]$) you can express the function $x[x+3]$ as a sum of 2 non-decreasing functions.
Now  $\int_1^2 f(x) d(x[x+3]) = \int_1^2 f(x) d(4x) + \int_1^2 f(x) d(x[x-1])= 4\int_1^2 f(x) dx + 2f(2)$.
In the first integral we used $(4x)'=4$, the second one is similar to the first part. 
In both cases we used R-S integral for a function having single jump. You can read more about it at this blog: Integrating Across a Jump. We have also used additivity in integrator, see another entry at the same blog.
Hint for question 2: Use the additivity in integrator again, in this way you can get two integrals, in one case the integrator is differentiable, in the second one it is a function having single jump. I guess question 3 can be handled in the similar way, although you'll have more jumps, not a single one.
