# Prove that $f \in R(F_s)$ on the interval $[a, b]$ and that $\int f dF_s = f(s)$

Let $$a and let $$f:[a, b] \to \mathbb{R}$$ be a bounded function that is continous at the point s. Define

$$F_s(x) = \begin{cases} 0, & \text{if a \le x \lt s} \\ 1, & \text{if s \le x \le b} \end{cases}$$

Prove that $$f \in R(F_s)$$ on the interval $$[a, b]$$ and that

$$\int fdF_s = f(s)$$.

My attempt to the proof:

Consider the partition,$$P$$ of the interval $$[a, b]$$ where $$a=x_o \le x_1 \le ... \le x_{i-1} \le x_i \le ... \le x_n = b$$.

Let $$M_i = supf(x) and m_i = inf f(x)$$ for $$x_{i-1} \le x \le x_i$$.

Given $$\epsilon \gt 0$$, and $$P$$. $$f\in R(F_s)$$ if $$U(P,f, F_s)-L(P,f,F_s) \lt \epsilon$$.

We know $$F(x)$$ is continuous at $$s \in [a,b]$$. Let there exist subinterval $$[x_{i-1}, x_i]$$ in $$[a, b]$$ such that $$s \in (x_{i-1}, x_i)$$.

Let $$F(x_i) - F(x_{i-1}) \lt \frac{\epsilon}{2mM}$$. Let $$M_i - m_i \lt 2M$$. Then $$\sum_{i=0}^{n}(M_i - m_i) \lt 2mM$$ because there are exactly $$m$$ terms in this sum, as there are exactly $$m$$ of the intervals in $$[x_{i-1}, x_i]$$.

$$U(P,f, F_s) - L(P,f, F_s) = (F(x_i)-F(x_{i-1}))\sum_{i=0}^{n}(M_i - m_i) \lt 2mM \frac{\epsilon}{2mM} = \epsilon$$.

Therefore, $$f \in R(F_s)$$.

For the second part of the proof, we need to show: $$\int fdF_s = f(s)$$.

Am I doing this proof right so far? could someone help me with the rest?

Everything looks good. You've proven that $$f$$ is Riemann-Lebesgue integrable with respect to $$F_s$$. Now pick any given sequence of partitions that allow the integral to converge $$\Pi_k$$ and focus on the partition points just before and after $$s$$. Their contribution is $$f(x_r)(F(x_r)-F(x_{r-1}))=f(x_r)\cdot 1$$. Now just use continuity of $$f$$ at $$s$$ and the definition of a sequence of partitions to conclude the integral converges to $$f(s)$$.