Sum of Functions of Bounded Variation is of Bounded Variation

Theorem
Let $a, b$ be real numbers with $a < b$.

Let $f, g : \closedint a b \to \R$ be functions of bounded variation.

Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the total variations of $f$ and $g$ respectively, on $\closedint a b$.

Then $f + g$ is of bounded variation on $\closedint a b$ with:


 * $\map {V_{f + g} } {\closedint a b} \le \map {V_f} {\closedint a b} + \map {V_g} {\closedint a b}$

where $\map {V_{f + g} } {\closedint a b}$ denotes the total variation of $f + g$ on $\closedint a b$.

Proof
For each finite subdivision $P$ of $\closedint a b$, write:


 * $P = \set {x_0, x_1, \ldots, x_n }$

with:


 * $a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$

Then:

Note that since $f$ and $g$ are of bounded variation, there exists $M, K \in \R$ such that:


 * $\map {V_f} {P ; \closedint a b} \le M$
 * $\map {V_g} {P ; \closedint a b} \le K$

for all finite subdivisions $P$ of $\closedint a b$.

We therefore have:


 * $\map {V_{f + g} } {P ; \closedint a b} \le M + K$

for all finite subdivisions $P$.

So $f + g$ is of bounded variation.

Note then that: