Problem
Suppose $f$ is continuous and $\phi$ is of bounded variation on $[a, b]$.
Show that the function $\psi(x)=\int_a^xfd\phi$ is of bounded variation on $[a, b]$.
If $g$ is continous on $[a, b]$, show that $\int_a^b gd\psi=\int_a^b gfd\phi$.
My proof
- Since $f$ is continuous $f$ is uniformly continuous on $[a, b]$. That is, $$\forall\varepsilon>0, \exists\delta\gt0 ~s.t.~ \left|f(x)-f(y)\right|\lt\varepsilon~~if~~\left| x-y\right| \lt \delta~~for~~any~~x, y\in[a, b]. $$
- Since $\phi$ is of bounded variation on $[a, b]$, we have, for constant $M_1$, $$ S_\Gamma[\phi; a, b]=\sum_{i=1}^m{|\phi(x_i)-\phi(x_{i-1})|} ~~~\le~~~ V[\phi; a, b]=\sup_\Gamma{S_\Gamma} ~~~\le~~~ M_1 ~~~\lt~~~ +\infty $$ for any partition $\Gamma=\left\{a=x_0\lt x_1 \lt \cdots \lt x_m=b\right\}$.
- Since $f$ is continuous, $f$ is bounded on $[a, b]$. That is, for constant $M_2$, $$\left| f \right| \le M_2 \lt +\infty$$
- $\cdots$
- $\cdots$
Is it okay if I considered that $f$ is uniformly continuous and bounded on a specific interval, when given the condition that $f$ is continuous?