# Absolute Continuity (Part One)

/There are two definitions of absolute continuity out there. One refers to an absolutely continuous *function* and the other to an absolutely continuous *measure*. Sometimes in mathematics, a single word or phrase gets recycled and adopts different meanings. Take *normal* for instance. It's sorely abused! A *normal* subgroup, a *normal* field extension, a *normal *vector to a surface, a *normal* topological space, a *normal *distribution. The list goes on. Oi!

This, however, is not the case with absolute continuity. Although the definitions of an **absolutely continuous function** and an **absolutely continuous measure** are different, they are intrinsically related, linked together by Lebesgue's Fundamental Theorem of Calculus:

*A function $F:[a,b]\to\mathbb{R}$ is absolutely continuous if and only if $$\int_a^xF'\;d\mu=F(x)-F(a) \qquad \text{for all $x\in[a,b]$}$$ where the integral is a Lebesgue integral.**

And an absolutely continuous *measure* is needed to prove one direction of this statement. So to put it succinctly,

*absolute continuity characterizes which functions can be an antiderivative.*

Below we'll give the formal definitions of absolute continuity and then discuss their relationship to this Fundamental Theorem of Lebesgue theory, leaving some of the details until next week.

On to the definitions:

**absolutely continuous**if for all $\epsilon>0$ and all $n>0$ there exists $\delta>0$ such that whenever $\{(a_i,b_i)\}_{i=1}^n\subset[a,b]$ is a pairwise disjoint collection of open intervals with $\sum_{i=1}^n b_i-a_i < \delta$, we have $\sum_{i=1}^n |f(b_i)-f(a_i)| < \epsilon$.

**absolutely continuous**with respect to $\mu$ (notationally, $\nu \ll \mu$) if $\mu(A)=0$ implies $\nu(A)=0$ for all $A\in\Sigma$.

*Let $f:[a,b]\to [0,\infty]$ be an integrable function and suppose $F(x)=\int_a^xf\;d\mu$. Then $F$ is absolutely continuous.*

To prove this, it's enough to verify the following lemma:

**Lemma**: For any measurable set $A\subset\mathbb{R}$ and any integrable function $f:[a,b]\to [0,\infty]$, for each $\epsilon>0$ there exists $\delta>0$ such that $$ \mu(A)<\delta \quad \text{ implies }\quad \int_A f\;d\mu < \epsilon. $$

To see why this is enough, assume for the moment the lemma is true. Let $\{(a_i,b_i)\}_{i=1}^n$ be a disjoint collection of open intervals in $[a,b]$ and set $A=\bigcup_{i=1}^n(a_i,b_i)$. Then for a fixed $\epsilon >0$ there is a $\delta >0$ so that $\mu(A)< \delta$ implies $\int_a^b f\;d\mu < \epsilon$. Therefore \begin{align*} \sum_{i=1}^n|F(b_i)-F(a_i)| &= \sum_{i=1}^n\int_{a_i}^{b_i}f\;d\mu \\ &=\int_{\cup_i(a_i,b_i)}f\;d\mu\\ &=\int_A f\;d\mu\\ & < \epsilon, \end{align*} which proves that $F$ is absolutely continuous.

So the real key is to show that the statement in the Lemma holds, and for this we'll use absolute continuity of measures! This can be done in three simple steps. I've outlined them below, and next week we'll fill in the details.

**Proof of the Lemma**

**Step 1**

Show that if $\nu\ll\mu$, then for all $\epsilon>0$ there is a $\delta>0$ such that $\mu(A)<\delta$ implies $\nu(A)<\epsilon$ for all measurable sets $A$.

**Step 2**

For any measurable set $A$, define $$\nu(A)=\int_Af\;d\mu$$ and show that $\nu$ is a measure.

**Step 3**

Observe that the $\nu$ defined in Step 2 satisfies $\nu\ll\mu$, thus we may apply Step 1 to conclude that for all $\epsilon>0$ there is a $\delta>0$ such that $\mu(A)< \delta$ implies $\nu(A)=\int_Af\;d\mu<\epsilon$.

*Real Analysis*(which I highly recommend!):

"The absolute continuity of $F(x)=\int_a^xF'$ can be regarded as a condition on the measure $\nu(A)=\int_A g$, namely $\nu(A)< \epsilon$ whenever $\mu(A)< \delta$, or $\nu(A)\to 0$ as $\mu(A)\to 0$. In this sense, absolute continuity is a continuity (proper) of certain measures."

-- *Real Analysis*, p. 371. *(notation changed to fit our discussion.) *

*Footnotes:*

* Actually, there's a bit more to say in the "if" direction. Precisely: if $F$ is absolutely continuous, then $F$ is differentiable almost every where, $F'$ is integrable, and $\int_a^x F'\;d\mu=F(x)-F(a)$. For the proof, see Royden's *Real Analysis* (4ed) ch. 6.5 Theorem 10, or Folland's *Real Analysis* (2ed) ch. 3.5, Theorem 3.35.

Briefly, it goes something like this: since $F$ is absolutely continuous, it is of bounded variation (Carothers, Proposition 20.15). Therefore $F$ can be written as the difference of two monotone functions (Royden, ch. 6.3 p. 117) and so it is differentiable almost everywhere (ibid., ch. 6.2 p. 112) and its derivative $F'$ is integrable and $\int_a^b F'=F(b)-F(a)$ (ibid., ch. 6.5 Theorem 10).