Inspired by Rishabh Deep Singh

I will attempt to do this problem "from first principles", going back to the definition, with the notations used here .

Consider an approximating sum $\sum_{j=0}^{n-1}f(c_j)\left(g(x_{j+1}-g(x_j)\right)$ with the integrand $f(x)=\lfloor x-0.5\rfloor$ and the integrator $g(x)=\lfloor x\rfloor$ , with a mesh size of less than 0.5, meaning that $x_{j+1}-x_j<0.5$ for all $j=0,..,n-1$ . Consider a positive integer $k$ with $1\leq k \leq 100$ , arbitrary but fixed. If $j$ is the largest index such that $x_{j}< k$ , then $c_j$ will be on the interval $(k-0.5,k+0.5)$ , so that $f(c_j)=k-1$ . Also, $g(x_j)=k-1$ and $g(x_{j+1})=k$ .Thus the integer $k$ makes a contribution of $f(c_j)\left(g(x_{j+1}-g(x_j)\right)=(k-1)1=k-1$ to the approximating sum. Adding up the contributions of all $k$ , we find that the approximating sum is $\sum_{k=1}^{100}(k-1)=\frac{99\times 100}{2}=4950$ .

Since all approximating sums with a mesh size of less than 0.5 are 4950, the Riemann-Stieltjes integral exists and has the same value. The answer is $\boxed{4950}$ .

We can phrase this solution a little more generally as follows: If $f(x)$ is continuous at the integers, then we have the Riemann-Stieltjes integral $\int_{a}^{b}f(x)d\lfloor x \rfloor =\sum f(c)$

where the sum is taken over all integers $c$ with $a<c\leq b$ . For example, $\int _{ 0 }^{ 2 }{ ( { x }^{ 2 }+1 ) \, d\lfloor x \rfloor } =f(1)+f(2)=2+5=7$

in the inspirational problem.

I'm going to use 3 propositions without proofs:

1.- Let's suppose that $f$ is a Riemann-Stieljes integrable function in $[a , c]$ and in $[c , b]$ ( a < c < b ) respect to function $g$ , then $f$ is a Riemann-Stieljes integrable function respect to $g$ in $[a , b]$ and $\int_a^b f \space dg = \int_a^c f \space dg + \int_c^b f \space dg$ .

This first proposition is only true if we are go to use (1) the generalizaded R-S integral definition here . It's not true if we are going to use the definition here , so the proof for this question will only be rigth if we are using (1) the generalized R-S integral . In other case is wrong...

For one proof of this proposition, take one partition P in [a,b] being the union of partitions P' in [a,c] and P'' in [c,b] given for $\frac{\epsilon}{2} > 0$ and use the triangular inequality for the absolute value....Briefly and gruffly, $|A + B - S(P,f,g)| \leq |A - S(P',f,g)| + |B - S(P'',f,g)| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ ...

2.- Let $f$ to be a real bounded function in $[a , b]$ right-continuous in $a$ and left-continuous in $b$ , and $g(x) = k$ a constant function in $(a , b)$ such that $g(a) \leq k \leq g(b)$ with $k$ a real constant, then $f$ is a Riemann-Stieljes integrable function respect to $g$ and $\int_a^b f \space dg = f(a) \cdot ( k - g(a)) + f(b) \cdot (g(b) - k)$

3.- If $f$ is a real bounded function and $a \in \mathbb{Z}$ then $\int_{a}^{a + 0.5} f \space d\lfloor x \rfloor = 0$ . Proof: use upper and lower Riemann-Stieljes sums...

Solution:

$\int_0^{100} \lfloor x - 0.5 \rfloor \space d\lfloor x \rfloor =$ $= \int_0^{0.5} \lfloor x - 0.5 \rfloor \space d\lfloor x \rfloor + \int_{0.5}^{1} \lfloor x - 0.5 \rfloor \space d\lfloor x \rfloor + \int_1^{1.5} \lfloor x - 0.5 \rfloor \space d\lfloor x \rfloor + ... + \int_{99}^{99.5} \lfloor x - 0.5 \rfloor \space d\lfloor x \rfloor + \int_{99.5}^{100} \lfloor x - 0.5 \rfloor \space d\lfloor x \rfloor =$ $\int_{1.5}^{2} \lfloor x - 0.5 \rfloor \space d\lfloor x \rfloor + \int_{2.5}^{3} \lfloor x - 0.5 \rfloor \space d\lfloor x \rfloor + ... + \int_{99.5}^{100} \lfloor x - 0.5 \rfloor \space d\lfloor x \rfloor =$ $= 1 + 2 + 3 + ... + 99 = \frac{(99 + 1)99}{2} = 4950$