如果$f\in R(\alpha)[a,b]$,则有$\alpha\in R(f)[a,b]$,而且$$\int_a^bfd\alpha+\int_a^b\alpha df=f(b)\alpha(b)-f(a)\alpha(a)$$.
现在用阿贝尔变换证明它：$f\in R(\alpha)[a,b]$说明存在实数$L$,使得对于任意给定的正实数$\varepsilon$,都存在$[a,b]$的分割$P$,使得对于$P$的任何加细$P'$,都有
$$|S(P',\alpha,f)-L|<\varepsilon$$
即
$$|\sum_{i=0}^{n-1}f(x_{ti})(\alpha(x_{i+1})-\alpha(x_i))-L|<\varepsilon$$
其中$x_{ti}$是在区间$[x_i,x_{i+1}]$上的任意一个数，$\{x_0,\cdots,x_n\}=P'$.根据阿贝尔变换,
$$\sum_{i=0}^{n-1}f(x_{ti})(\alpha(x_{i+1})-\alpha(x_i))=f(x_{t,n-1})[\alpha(x_n)-\alpha(x_0)]-\sum_{i=0}^{n-2}(\alpha(x_{i+1})-\alpha(x_0))(f(x_{t,i+1})-f(x_{ti}))$$
即
$$\sum_{i=0}^{n-1}f(x_{ti})(\alpha(x_{i+1})-\alpha(x_i))=f(x_{t,n-1})[\alpha(x_n)-\alpha(x_0)]-\sum_{i=0}^{n-2}\alpha(x_{i+1})(f(x_{t,i+1})-f(x_{ti}))+\alpha(x_0)(f(x_{t,n-1})-f(x_{t0}))$$
即
$$\sum_{i=0}^{n-1}f(x_{ti})(\alpha(x_{i+1})-\alpha(x_i))=f(x_{t,n-1})[\alpha(b)-\alpha(a)]-\sum_{i=0}^{n-2}\alpha(x_{i+1})(f(x_{t,i+1})-f(x_{ti}))+\alpha(a)(f(x_{t,n-1})-f(x_{t0}))$$
我们令$x_{t,n-1}=b$,$x_{t0}=a$,(为什么可以这样令?).则
$$\sum_{i=0}^{n-1}f(x_{ti})(\alpha(x_{i+1})-\alpha(x_i))+\sum_{i=0}^{n-2}\alpha(x_{i+1})(f(x_{t,i+1})-f(x_{ti}))=f(b)(\alpha(b)-\alpha(a))+\alpha(a)(f(b)-f(a))=f(b)\alpha(b)-f(a)\alpha(a)$$
我们很高兴地发现，$\{x_{t0},\cdots,x_{t,n-1}\}$是$[a,b]$的一个分割$Q'$,而且$x_{t,i-1}\leq x_i\leq x_{t,i}$.因此
$$\int_a^bfd\alpha+\int_a^b\alpha df=f(b)\alpha(b)-f(a)\alpha(a)$$(为什么?)