Bounded variation, difference of two increasing functions

Solution 1:

• Let $f$ a function of bounded variation. Let $F(x):=\sup \sum_{j=1}^{n-1}|f(x_{j+1})-f(x_j)|=:\operatorname{Var}[a,x]$, where the supremum is taken over the $x_1,\ldots,x_n$ which satisfy $a=x_1<x_2<\ldots<x_n=x$. Since $f$ is of bounded variation, $F$ is bounded, and by definition increasing. Let $G:=F-f$. We have to show that $G$ is bounded and increasing. Boundedness follows from this property for $f$ and $F$, now fix $a\leq x_1<x_2\leq b$. We have $$G(x_2)-G(x_1)=F(x_2)-f(x_2)-F(x_1)+f(x_1)\geq 0$$ because $\operatorname{Var}[a,x_1]+f(x_2)-f(x_1)\leq \operatorname{Var}[a,x_1]+|f(x_2)-f(x_1)|\leq \operatorname{Var}[a,x_2]$.
• If $f$ and $g$ are of bounded variation so is $f-g$. If $f$ is increasing then we have, if $a=x_0<x_1<\ldots<x_n=b$ that $\sum_{j=1}^{n-1}|f(x_{j+1})-f(x_j)|=|f(b)-f(a)|$, so $f$ is of bounded variation. So the difference of two bounded monotonic increasing functions is of bounded variation.