Absolutely continuous functions

Note

A function $f:[a,b]\to \mathbb{R}$ is called absolutely continuous if for every $ϵ>0$ there exists ${\delta }_{ϵ}>0$ such that every finite familiy of open disjoint intervals $(a_k,b_k)$ such that the sum of their lenghts is less than ${\delta }_{ϵ}$

$$\sum _{k=1}^n(b_k-a_k)<{\delta }_{ϵ}$$

the inequality

$$\sum _{k=1}^n|f(b_k)-f(a_k)|<ϵ$$

is satisfied.

Examples §

Example

The function $\sqrt{x}\in AC[0,1]$. Note that this function is not uniformly continuous, so that $UC\text{centernot}⟹AC$. We use the fact that $0\le h\le a<b$

$$\sqrt{b}-\sqrt{a}\le \sqrt{b-h}-\sqrt{a-h}$$

we can then shift all intervals to the left, concatenating them to obtain a telescopic sum

$$\sum _k(\sqrt{b_k}-\sqrt{a_k})\le \sqrt{c_n}$$

where $c_1=0$ and $c_2=b_1-a_1$, and so on. Then $c_n$ is the total lenght of the intervals, just choose $\delta ={ϵ}^2$.

Properties §

It’s a stronger condition than uniform continuity, just take a family of one interval. It’s not equivalent, since the Cantor-Vitali function, which is continuous, hence in $[0,1]$ is uniformly continuous is not AC since we can always find a family $(a_k,b_k)$ such that

$$\sum _{k=1}^n|c(b_k)-c(a_k)|=1,\sum _{k=1}^n(b_k-a_k)<\delta$$

the “rise” of the function is on a set of measure zero!

Clealry if $f$ is Lipschitz, then it’s absolutely continuous, since

$$\sum _{k=1}^n|f(b_k)-f(a_k)|\le L\sum _{k=1}^n(b_k-a_k)$$

just take $\delta =\frac{ϵ}{L}$.

We introduce this definition in our quest to find the family of functions for which the fondamental theorem of calculus holds, hence it’s not surprising that given any $f\in L^1(a,b)$, then the integral function

$$F(x):={\int }_a^{x}fdx$$

it’s absolutely continuous. This follows from the absolutely continuity of the Lebesgue integral (hence the name).

Let $(a_k,b_k)$ a famili of disjoint open intervals. Then

$$\sum _k|F(b_k)-F(a_k)|=\sum _k\left|{\int }_{a_k}^{b_k}fdx\right|\le \sum _k{\int }_{a_k}^{b_k}|f|dx={\int }_{{\cup }_k(a_k,b_k)}|f|dx<ϵ$$

if the measure of the set we are integraitng on is small enough:

$$\mu \left(\bigcup _k(a_k,b_k)\right)=\sum _k(b_k-a_k)<{\delta }_{ϵ}$$

since the intervals are disjoint.

The following theorem makes clear that the family of function that satisfies the Fundamental Theorem of Calculus is precisley $AC$.

Note

Let $f\in C^0[a,b]$. $f$ is $AC[a,b]$ iff $\{{\Delta }_{h}f{\}}_h$ is uniformly-integrable.

Note \{\Delta_h f\}_h is uniformly-integrable¹, then every integral function shares the same continuity modulus \delta_\epsilon. Let (a_k, b_k) be a family of disjoint interval with total length less than \delta_\epsilon

Suppose that

$$\sum _{k=1}^n\left|\frac{1}{h}{\int }_{b_k}^{b_k+h}fdt-\frac{1}{h}{\int }_{a_k}^{a_k+h}fdt\right|=\sum _{k=1}^n\left|{\int }_{a_k}^{b_k}{\Delta }_{h}fdt\right|<ϵ$$

Taking the limit $h\to 0$, and using continuity of $f$ we get that $f\in AC[a,b]$. Now suppose $f\in AC[a,b]$, WLOG we can assume that $f$ is non-decreasing (the usual Jordan decomposition), then ${\Delta }_{h}f\ge 0$. Let $A={\bigcup }_k(a_k,b_k)$ the usual disjoint union of intervals. Then

$${\int }_A{\Delta }_{h}fdt=\sum _k{\int }_{a_k}^{b_k}{\Delta }_{h}fdt=\frac{1}{h}\sum _k{\int }_{b_k}^{b_k+h}fdt-{\int }_{a_k}^{a_k+h}fdt$$ $$=\frac{1}{h}{\int }_0^h\sum _{k}f(b_k+t)-f(a_k+t)dt<\frac{1}{h}{\int }_0^hϵdt=ϵ\square$$

Note

Let $f\in AC([a,b])$. Then its derivative $f^{\prime }$ exists a.e., furthermore $f^{\prime }\in L^1(a,b)$ and satisfies:

$${\int }_a^x{f}^{\prime }dt=f(x)-f(a),\forall x\in [a,b]$$

Note AC \implies BV, f can be decomposed as the difference of two monotonic functions, which have derivatives a.e. For the previous lemma, and the fact that \Delta_h f \to f', using Vitali's theorem:

Since

$${\int }_a^x{f}^{\prime }dt=\underset{h\to 0}{\lim }{\int }_a^x{\Delta }_{h}fdt=\underset{h\to 0}{\lim }\frac{1}{h}\left({\int }_x^{x+h}fdt-{\int }_a^{a+h}fdt\right)=f(x)-f(a)\square$$