Absolutely Continuous Functions are of Bounded Variation

# Absolutely Continuous Functions are of Bounded Variation

Recall from the Absolute Continuity page that a function $f : [a, b] \to \mathbb{R}$ is said to be absolutely continuous on $[a, b]$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all finite collections of disjoint open subintervals of $[a, b]$, $\{ (x_1, y_1), (x_2, y_2), ..., (x_n, y_n) \}$ with $\displaystyle{\sum_{i=1}^{n} (y_i - x_i)}$ we have that:

(1)\begin{align} \quad \sum_{i=1}^{n} | f(y_i) - f(x_i) | < \epsilon \end{align}

We noted that every absolutely continuous function $f$ on $[a, b]$ is uniformly continuous on $[a, b]$ and hence continuous on $[a, b]$.

We now show that every absolutely continuous function $f$ on $[a, b]$ is of bounded variation on $[a, b]$.

Theorem 1: If $f : [a, b] \to \mathbb{R}$ is absolutely continuous on $[a, b]$ then $f$ is of bounded variation on $[a, b]$. |

- Since $f$ is absolutely continuous on $[a, b]$ for $\epsilon = 1 > 0$ there exists a $\delta > 0$ such that for every finite collection of disjoint open subintervals of $[a, b]$, $\{ (x_1, y_1), (x_2, y_2), ..., (x_n, y_n) \}$ with $\displaystyle{\sum_{i=1}^{n} (y_i - x_i) < \delta}$ we have that:

\begin{align} \quad \sum_{i=1}^{n} |f(y_i) - f(x_i)| < \epsilon = 1 \end{align}

- Let $P^* = \{ a = a_0, a_1, ..., a_n = b \}$ be a partition of $[a, b]$ with the property that $a_i - a_{i-1} = \frac{\delta}{2}$ for all $i \in \{ 1, 2, ..., n - 1 \}$, and such that $a_n - a_{n-1} \leq \frac{\delta}{2}$.

- Then we have that:

\begin{align} \quad n = \left \lfloor \frac{2(b - a)}{\delta} \right \rfloor + 1 \end{align}

- (This is because the length of the interval $(a, b)$ is $b - a$. We then divide this length by $\displaystyle{\frac{\delta}{2}}$ and take the floor of this number. This gives us the number of times $\displaystyle{\frac{\delta}{2}}$ goes into $b - a$. We add $+1$ to account for the remaining interval $(a_{n-1}, a_n)$ whose less is less than or equal to $\displaystyle{\frac{\delta}{2}}$.)

- Now let $P$ be any partition of $[a, b]$ and let:

\begin{align} \quad P' = P \cup P^* \end{align}

- Then $P'$ is a refinement of $P$ and $P^*$. Let $P' = \{ z_0 = a, z_1, ..., z_m = n \}$. For each $i \in \{ 1, 2, ..., n \}$ let:

\begin{align} \quad P_i' = \{ z_{i_k} \in P' : z_{i_k} \in [a_{i-1}, a_i] \} \end{align}

- That is, $P_i'$ is the sets of points in $P_i$ that are contained in the closed interval $[a_{i-1}, a_i]$ from the partition $P^*$. Then:

\begin{align} \quad V(P, f) \leq V(P', f) = \sum_{i=1}^{n} \sum_{k} | f(z_{i_k}) - f(z_{i_k - 1}) | \leq \sum_{k=1}^{n} 1 = n = \left \lfloor \frac{2(b - a)}{\delta} \right \rfloor + 1 \end{align}

- Therefore the variation of $V(P, f)$ is always bounded by $\displaystyle{\left \lfloor \frac{2(b - a)}{\delta} \right \rfloor + 1}$, so $f$ is of bounded variation on $[a, b]$.