Ex2-16

Chapter 2 Question 16

First we show a special case where $$f$$ is continuous except for a jump discontinuity at $$a$$,. We define $$\tilde{f}$$ to be the continuous function that agrees with $$f$$ on $$(a,b]$$ but

$$\tilde{f}(a)=\lim_{x\rightarrow a^+}f(x)$$

so $$\tilde{f}$$ is continuous on $$[a,b]$$. Then we claim that $$\int_a^bfd\phi$$ exists and

$$\int_a^bfd\phi=\int_a^b\tilde{f}d\phi$$

Let $$|f(x)|\leq M$$ for all $$x, \phi$$ s continuous at $$a$$. Given $$\epsilon>0$$, choose $$\delta>0$$ such that

$$|\phi(x)-\phi(a)|<\epsilon\quad\text{ if } |x-a|<\delta$$

Then for any partition $$\Gamma$$ with $$|\Gamma|<\delta$$, choosing the same $$\xi_i$$'s for $$f,\tilde f$$ except possibly the one in $$[a,x_1]$$, we have

$$\begin{align}


 * S_{\Gamma}(f)-S_{\Gamma}(\tilde{f})|&=\left|\left(f(\xi)-\tilde{f}(\xi')\right)\left(\phi(x_1)-\phi(a)\right)\right|\\

&<2M\epsilon

\end{align}$$

By taking $$\epsilon$$ arbitrarily small and noting that $$\lim_{|\Gamma|\rightarrow0}S_{\Gamma}(\tilde{f}

)$$ exists, we conclude

$$\lim_{|\Gamma|\rightarrow0}S_{\Gamma}\left(f\right)=\int_a^bfd\phi=\int_a^b\tilde{f}d\phi$$

The case where $$f$$ has a jump discontinuity at $$b$$ or at both of $$a$$ and $$b$$ is an easy extension of the argument above.

We return to the original problem. We can assume $$\phi$$ is increasing on $$[a,b]$$ to prove the statement. Note that for a general $$\phi$$ of bounded variation, we can write it as $$\phi=\phi_1-\phi_2$$, the difference of two increasing functions. Assume $$|f(x)|\leq M$$for all $$x$$, and suppose $$f$$ has jump discontinuities at $$x_1,\cdots,x_n$$, then define $$f_0,f_1,\cdots,f_n$$ on $$[a,x_1],[x_1,x_2],\cdots,[x_n,b]$$ such that

$$f_i(x)=f(x)\quad\text{ for }x\in[x_i,x_{i+1}]$$

where $$a=x_0,b=x_{n+1}$$. Now we claim $$\int_a^bfd\phi$$ exists and

$$\int_a^bfd\phi=\sum_{i=0}^n\int_{x_i}^{x_{i+1}}f_id\phi$$

Given $$\epsilon>0$$, choose $$\delta_1>0$$ such that for partition $$\Gamma_i$$ of $$[x_i,x_{i+1}]$$ with $$|\Gamma_i|<\delta_1$$,

$$\left|S_{\Gamma_i}(f_i)-\int_{x_i}^{x_{i+1}}f_id\phi\right|<\epsilon$$

Let $$D=\{x_1,x_2,\cdots,x_n\}$$ and let $$\phi$$ be continuous at each point of $$D$$. Choose $$\delta_2>0$$ such that for each $$i=1,2,\cdots,n$$

$$|\phi(x)-\phi(y)|<\epsilon\quad\text{ if }x,y\in(x_i-\delta_2,x_i+\delta_2)$$

we also assume $$\delta_2$$ to be small enough such that each $$(x_i-\delta_2,x_i+\delta_2)$$ are disjoint.

Let $$\delta=\min(\delta_1,\delta_2)$$, for partition $$\Gamma$$ of $$[a,b]$$ with $$|\Gamma|<\delta$$, we let $$\Gamma\cup D$$ be the refinement with points $$\{x_1,\cdots,x_n\}$$, then since $$\phi$$ is increasing,

$$S_{\Gamma\cup D}(f)\geq S_{\Gamma}(f).$$

We can choose two sets of $$\{\xi_i\}, \{\xi_i'\}$$ in the Riemann-Stieltjes sum such that $$\xi_i=\xi_i'$$ on intervals of $$\Gamma$$ which don't contain any points from $$D$$. Then if we define $$M=\sup_{x\in[a,b]} |f(x)| $$

$$\begin{align}&\left|S_{\Gamma\cup D}(f)- S_{\Gamma}(f)\right| \\&\leq \sum_{i=1}^n ((\sup_{x\in[a_i,b_i]}|f(x)|)|\phi(x_i)-\phi(a_i)|\\ &+(\sup_{x\in[a_i,b_i]}|f(x)|)|\phi(b_i)-\phi(x_i)|+(\sup_{x\in[a_i,b_i]}|f(x)|)|\phi(b_i)-\phi(a_i)|)\\ &\leq n\cdot 3M \cdot\epsilon\end{align}$$

where $$[a_i,b_i]$$is a subinterval in $$\Gamma$$ containing $$x_i$$ for all $$i=1,2,\ldots,n$$.

Let $$\Gamma_0, \Gamma_1, \cdots, \Gamma_n$$ be partitions respectively of $$[a_0,b_0], [a_1,b_1],\ldots, [a_n,b_n].$$ The union of those partitions is $$\Gamma\cup D$$, and then

$$S_{\Gamma\cup D}(f)=S_{\Gamma_0}(f_0)+\cdots+S_{\Gamma_n}(f_n)$$

implies

$$\left|S_{\Gamma \cup D}(f)-\sum_{i=0}^n\int_{x_i}^{x_{i+1}}f_id\phi\right|<(n+1)\epsilon$$ implies

$$\left|S_{\Gamma}(f)-\sum_{i=0}^n\int_{x_i}^{x_{i+1}}f_id\phi\right|<(3nM+n+1)\epsilon$$

choose $$\epsilon$$ small enough, and we have completed the proof.