Show that if $f,g \in BV([a,b]) $ then $$fg \in BV([a,b]) $$ My idea was to write $f = f_1 - f_2$ and $g = g_1 - g_2$ (which are non decreasing functions) and write $fg$ as: $$ fg = (f_1 g_1 + f_2 g_2 ) - (f_1 g_2 + f_2 g_1)$$
which are two non decreasing function hence $fg$ is a bounded variation function. Is this correct?
Proposition: The difference of two increasing functions on a compact set is BV
Let $f_1,f_2$ be increasing and $g=f_1-f_2$. Now consider any partition of of $[a,b]$, $a=x_0<x_1<...<x_n=b$.
$$\sum^n_{j=1}|g(x_j)-g(x_{j-1})| \\=\sum|f_1(x_j)-f_2(x_j)-f_1(x_{j-1})+f_2(x_{j-1})|\\=\sum |(f_1(x_j)-f_1(x_{j-1}))+(-f_2(x_j)+f_2(x_{j-1}))|\\ \le \sum |f_1(x_j)-f_1(x_{j-1})|+|f_2(x_j)-f_2(x_{j-1})|\\=\sum f_1(x_j)-f_1(x_{j-1})+f_2(x_j)-f_2(x_{j-1})\\=f_1(x_n)-f_1(x_0)+f_2(x_n)-f_2(x_0)\\\square$$
EDIT: The question has changed since I looked, but yes, what you have now is correct and the above proves it.