Function of Bounded Variation is Bounded

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

Let $f : \closedint a b \to \R$ be a function of bounded variation.

Then $f$ is bounded.

Proof
We use the notation from the definition of bounded variation.

Since $f$ is of bounded variation, there exists $M \in \R$ such that:


 * $\map {V_f} P \le M$

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

Let $x$ be a real number with:


 * $a < x < b$

Then $\set {a, x, b}$ is a finite subdivision of $\closedint a b$.

We have:


 * $\map {V_f} {\set {a, x, b} } = \size {\map f x - \map f a} + \size {\map f b - \map f x}$

Since $x \in \openint a b$ was arbitrary, we therefore have:


 * $\size {\map f x - \map f a} + \size {\map f b - \map f x} \le M$

for all $x \in \openint a b$.

We have:

So for all $x \in \openint a b$, we have:


 * $\size {\map f x} \le \size {\map f a} + M$

Since $M \ge 0$, this inequality is also satisfied for $x = a$.

We therefore have:


 * $\size {\map f x} \le \map \max {\size {\map f a} + M, \size {\map f b} }$

for all $x \in \closedint a b$.

So $f$ is is bounded.