# Absolute Continuity (Part Two)

In last week's post, we saw how absolute continuity - of a function and of a measure - is key in any discussion of Lebesgue's Fundamental Theorem of Calculus. In particular, these two types of absolute continuity appear in the proof that

#### if $f:[a,b]\to[0,\infty]$ is an integrable function and  $F(x)=\int_a^xf\;d\mu$, then $F$ is absolutely continuous.

The bulk of the proof (and where we see aboslute continuity in action) stems from the following

#### Lemma: For any (Lebesgue) measurable set $A$ and for all $\epsilon>0$ there is a $\delta>0$ such that $\mu(A)<\delta$ implies $\int_Af\;d\mu< \epsilon$.

Last time we showed why this is enough to imply the absolute continuity of $F$, and we closed with a brief three-step outline of the lemma's proof. Our goal for today is to to fill in the details.

## Proving the Lemma

### Step 1

Claim: Let $\nu$ and $\mu$ be positive measures, with $\nu$ finite*, on a measure space $(X,\Sigma)$. If $\nu$ is absolutely continuous with respect to $\mu$ (i.e. $\nu\ll\mu$), then for all $\epsilon>0$ there exists $\delta>0$ such that $\mu(A)< \delta$ implies $\nu(A)< \epsilon$ for all measurable sets $A$.

Proof: Suppose $\nu \ll \mu$, i.e. $\nu(A)=0$ whenever $\mu(A)=0$ for any measurable set $A$. But suppose to the contrary that there exists an $\epsilon>0$ such that for all $n\in\mathbb{N}$,    $$\mu(A_n)<\frac{1}{2^n} \quad \text{yet} \quad \nu(A_n)\geq \epsilon \quad \text{for each n.}$$  For each $k\in\mathbb{N}$, define $$F_k=\bigcup_{n=k}^\infty A_n$$ and observe that $F_1\supset F_2\supset F_3\supset \cdots$. Thus by continuity from above** \begin{align*} \mu\left( \bigcap_{k=1}^\infty F_k  \right) &= \lim_{k\to\infty}\mu (F_k)\\  &= \lim_{k\to\infty}\mu \left( \bigcup_{n=k}^\infty A_n  \right)\\  &\leq \lim_{k\to\infty} \sum_{n=k}^\infty \mu(A_n)\\  &=\lim_{k\to\infty} \frac{1}{2^{k-1}}\\  &=0. \end{align*} On the other hand, since $\nu(A_n)\geq \epsilon$ for all $n$, it follows that for every $k$,  $$\nu(F_k)=\nu\left( \bigcup_{n=k}^\infty A_n \right)\geq \epsilon$$  and so again by continuity from above,  $$\nu\left( \bigcap_{k=1}^\infty F_k\right)=\lim_{k\to\infty}\nu(F_k)\geq \epsilon.$$ Thus, if we let $A=\bigcap_{k=1}^\infty \bigcup_{n=k}^\infty A_n$ we see that $\mu(A)=0$ yet $\nu(A)\geq \epsilon$, contradicting the fact that $\nu \ll \mu$.

### Step 2

Claim: Suppose $\nu$ and $\mu$ are positive measures on a measure space $(X,\Sigma)$. For any measurable set $A$ and integrable function $f:X\to[0,\infty]$, define $$\nu(A)=\int_Af\;d\mu.$$ Then $\nu$ is a measure.

Proof: To verify that $\nu$ is a measure, we have to check three things:

• That $\nu(A)\geq 0$ for all measurable sets $A$ is clear since we're assuming $f$ itself is a positve function.***
• Since $\mu$ is a measure, we know $\mu(\varnothing)=0$. Thus $\nu(\varnothing)=\int_{\varnothing}f\;d\mu=0$ too. (In other words, $\nu \ll \mu$.)
• Suppose $\{A_n\}$ is a countable collection of pairwise disjoint measurable sets. Then    \begin{align*}    \nu\left(\bigcup_{n=1}^\infty A_n\right) &= \int_{\cup_n A_n}f\;d\mu\\      &=\sum_{n=1}^\infty\int_{a_n} f\;d\mu\\      &=\sum_{n=1}^\infty\nu(A_n)    \end{align*}  which shows that $\nu$ is countable additive.

### Step 3

Claim: The Lemma is proved!

Proof: Let $\nu$ and $\mu$ be defined as in Step 2 and observe that $\nu$ is absolutely continuous with respect to $\mu$. Thus by Step 1, for all $\epsilon>0$ there is a $\delta>0$ such that $\mu(A) < \delta$ implies $\nu(A)=\int_Af\;d\mu < \epsilon$ for all measurable sets $A$. This is precisely the statement of the Lemma.

Footnotes:

*[Edited 3/2/16] We need $\nu$ to be a finite measure! Many thanks to a reader for pointing this out in the comments.

**Notice that $\mu(F_1)<\infty$ since  \begin{align*}  \mu(F_1)&=\mu\left(\bigcup_{n=1}^\infty A_n  \right)\\  &\leq \sum_{n=1}^\infty \mu(A_n)\\  &=1.  \end{align*}

*** If $f$ is allowed to take on negative values, the math still holds, but we would need to work with a signed measure instead.

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