Ex2-31

Chapter 2, Question 31

Take $$\phi(x)=1$$ on $$[a,b]$$, since $$\phi$$ is constant, $$\int_a^bfd\phi$$ exists and

$$\int_a^bfd1=0$$

By theorem 2.21, $$\int_a^b1df$$ exists and

$$\int_a^b1df=f(b)\phi(b)-f(a)\phi(a)-\int_a^bfd1=f(b)-f(a)$$

Suppose $$f'$$ exists and is Riemann-integrable on $$[a,b]$$, then let $$I=\int_a^b f'dx$$ and $$J=f(b)-f(a)=\int_a^b df$$, we want to prove $$I=J$$. By definition of Riemann-integrable, given $$\varepsilon>0$$ we can choose $$\delta>0$$ so that for each partition with $$|\Gamma|<\delta$$and any choice of $$\xi_i\in[x_{i-1},x_i]$$, we have

$$|R_\Gamma - I |<\varepsilon\quad\text{ where } \quad R_\Gamma=\sum_{i=1}^k f'(\xi_i)(x_i-x_{i-1}).$$

Note that in fact, if we let $$S_\Gamma$$ denote the Riemann-Stieltjes sum, then

$$S_\Gamma=\sum_{i=1}^k 1\cdot (f(x_i)-f(x_{i-1}))=f(b)-f(a),$$

therefore, we know that $$S_\Gamma=J$$ for every partition $$\Gamma$$. By mean value theorem, there are some $$\xi_i\in[x_{i-1},x_i]$$ for $$i=1,2,\ldots,k$$ such that

$$f(x_i)-f(x_{i-1})=f'(\xi_i)(x_i-x_{i-1}).$$

Moreover, by restricting $$|\Gamma|<\delta$$ with these choices of $$\xi_i$$, we have $$R_\Gamma=S_\Gamma=J$$, thus $$|I-J|<\varepsilon$$. By choosing $$\varepsilon$$ arbitrarily small, we proved $$I=J$$.