Let $a<s<b$ 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?

Thanks in advance.


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 $f(s)$.


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