Prove that $f(0+)=f(0)$ if $f \in R(\beta_1)$ [closed]

Question

Let $\beta_1$ be a function defined by $$\beta_1(x)= \begin{cases} 0 & x \le 0\\ 1 & x >0 \end{cases} $$

Now we define $f(x)$ which is a bounded function on $[-1,1]$.

We need to how that $ f \in R(\beta_1)$ iff $f(0+)=f(0)$

Let us consider a partition $P_1$ on $[-1,\epsilon]$ then it is easy to see that $U(P_1,f,\beta_1)=0$ amd $L(P_1,f,\beta_1)=0$

Similarly on any partition $P_2$ on $[\epsilon,1]$ we see that $U(P_2,f,\beta_1)=0$ and $L(P_2,f,\beta_1)=0$.

Now, consider $[-\epsilon,\epsilon]$

$U(P_1,f,\beta_1)=\sup f(x)_{x \in [-\epsilon,\epsilon]}(\beta_1(\epsilon) - \beta_1(-\epsilon) = \sup(f(x))_{x \in [-\epsilon,\epsilon]}(-1) $ amd

$L(P_1,f,\beta_1)= \inf f(x)_{x \in [-\epsilon,\epsilon]}(\beta_1(\epsilon) - \beta_1(-\epsilon)=\inf(f(x))_{x \in [-\epsilon,\epsilon]}(1)$.

Then, by definition of Riemann-Stieltjes integral we see that,

$\sup(f(x))-\inf f(x))_{x \in [-\epsilon,\epsilon]}(1)< \epsilon$

Then $f(0) \le \sup(f(x))$ and also $-f(x+) \le -\inf f(x)$

Will I not be able to conclude that $\lim_{x \to 0+}f(x)=f(0)$?

Answer 1

Let $b:=\beta_1$. Let $R(b)$ denote the set of all Riemann–Stieltjes-integrable functions on $[-1,1]$ with respect to $b$. Then

$f\in R(b)$ iff $f$ is continuous at $0$.

Indeed, suppose that $f$ is continuous at $0$. Take any "partition" $P=(x_0,\dots,x_n)$ of $[-1,1]$ such that $-1=x_0<\dots<x_{k-1}\le0<x_k<\dots<x_n=1$ for some natural $k=k(P)\le n$. The corresponding upper and lower Riemann–Stieltjes sums for $f$ are $$U(P,f,b)=\sup_{[x_{k-1},x_k]}f,\quad L(P,f,b)=\inf_{[x_{k-1},x_k]}f.$$ If the mesh $$m(P):=\max_{j=1}^n(x_j-x_{j-1})$$ goes to $0$, then $x_k-x_{k-1}\to0$, and hence $x_{k-1}\to0$ and $x_k\to0$, so that $U(P,f,b)=\sup_{[x_{k-1},x_k]}f\to f(0)$ and $L(P,f,b)=\inf_{[x_{k-1},x_k]}f\to f(0)$, which implies $U(P,f,b)-L(P,f,b)\to0$. Thus, $f\in R(b)$ if $f$ is continuous at $0$.

Vice versa, suppose that $f\in R(b)$. Take any "partition" $P_n=(x_0,\dots,x_{2n})$ of $[-1,1]$ of mesh $m(P_n)=2/n$ such that $x_{n-1}=-1/n$ and $x_n=1/n$. Then $$\sup_{[-1/n,1/n]}|f-f(0)|\le U(P_n,f,b)-L(P_n,f,b)\to0.$$ Thus, $f$ is continuous at $0$ if $f\in R(b)$. $\qquad\Box$