Measure Theory/Absolute Continuity

Absolute Continuity

In light of the previous lesson, the following definition makes sense. Moreover, we hope that the Devil's staircase would not have this property (as that might explain why it fails the equality that we are hoping to ensure, $\int _{(a,b)}f'=f(b)-f(a)$).

Definition: absolute continuity

Let $f:[a,b]\to {\mathbb {R}}$ be any function. We say that f is absolutely continuous on [a,b] if it satisfies the following property.

$\forall \varepsilon \in {\mathbb {R}}^{+},\exists \delta \in {\mathbb {R}}^{+}$ such that for all collections of disjoint intervals with combined length less than $\delta$,

${\begin{aligned}&\{I_{1}=(a_{1},b_{1}),I_{2}=(a_{2},b_{2}),\dots ,I_{n}=(a_{n},b_{n})\},\\&\sum _{k=1}^{n}\ell (I_{k})=\sum _{k=1}^{n}(b_{k}-a_{k})<\delta \end{aligned}}$

we then have that the net change in f on these intervals is less than $\varepsilon$.

$\sum _{k=1}^{n}|f(b_{k})-f(a_{k})|<\varepsilon$

Exercise 1. Lipschitz to AC to UC Show that every Lipschitz continuous function is absolutely continuous, and that every absolutely continuous function is uniformly continuous.

AC to BV

If we are to hope that absolute continuity implies the desired equality $\int _{(a,b)}f'=f(b)-f(a)$ then of course we must be ensured that f' exists, or at least exists a.e.

Given our earlier studies, it might be especially elegant if we can simply show that absolute continuity implies bounded variance. Since we've shown that functions of bounded variance are differentiable a.e., then the differentiability of f a.e. follows immediately.

Exercise 2. AC Implies BV Let $f:[a,b]\to {\mathbb {R}}$ be any absolutely continuous function on a compact interval. To prove that it has bounded variation, we will pick a $\varepsilon \in {\mathbb {R}}^{+}$ and somehow use its corresponding $\delta$ due to absolute continuity. (I mean, how else would you use the absolute continuity property, right?) It will turn out not to matter which $\varepsilon$ we choose, so for convenience, set $\varepsilon =1$ and let $\delta$ be the response. 1. Consider any partition P. Let Q be the partition $Q=\{a,a+\delta ,a+2\delta ,\dots ,a+N\delta \}$ where N is the largest integer such that $a+N\delta \leq b$ Define the refinement of P by $P'=P\cup Q$ Show that if A is any partition and B is any refinement, $A\subseteq B$, then $V_{f}(A)\leq V_{f}(B)$ Infer that in our particular situation, $V_{f}(P)\leq V_{f}(P')$. 2. Let $P'=\{x_{0}=a<x_{1}<x_{2}<\cdots <x_{n}=b\}$ and split this partition into several smaller partitions, with end-points at $a,a+\delta ,a+2\delta ,\dots ,a+N\delta ,b$. Write $V_{f}(P')$ as a sum defined in terms of the points in P', and then split the sum into the smaller partitions described above. Argue that each component sum is less than 1. (If you find that, in fact, there is a bit of a logical problem at this point -- like maybe you wish that the smaller partitions had width strictly less than $\delta$, rather than having width exactly $\delta$, then go back and adjust the definition of Q to make things work out better.) 3. Use the above to show that $V_{f}(P')\leq N+1$. 4. Conclude the desired result.

AC to FTC

Finally we have what is necessary to state and prove the following conjecture: If a function, $f:[a,b]\to {\mathbb {R}}$ is absolutely continuous then $\int _{(a,b)}f'=f(b)-f(a)$.

Note that, due to theorems proved for functions of bounded variation, we already know that $\int _{(a,b)}f'$ exists and is bounded above by $f(b)-f(a)$.

Therefore if we define this area function,

$G(x)=\int _{(a,x)}f'$

it is well-defined. We already know, from the derivative of integrals, that then

$G'(x)=f'(x)$

almost everywhere.

We would like to infer that G and f are equal a.e. up to adding a constant term. But how?

We can simplify matters by not having two derivatives, but only one. The equation above is equivalent to $(G-f)'(x)=0$.

If we could then prove a zero derivative implies a constant function, we could quickly prove the desired result.

Exercise 2. Zero Derivative Let $f:[a,b]\to {\mathbb {R}}$ be an absolutely continuous function with $f'=0$ a.e. Prove that therefore f is constant. Below I'll give guiding steps to show that $f(b)=f(a)$. First let E be the set of points where $f'(x)=0$, and let $\varepsilon \in {\mathbb {R}}^{+}$. (At the end of the proof we will show that $|f(b)-f(a)|$ is small in terms of $\varepsilon$.) To each $x\in E$ associate with it a corresponding $y_{x}\in (x,b]$ such that for all $x<y<y_{x}$ we have ${\frac {|f(y)-f(x)|}{y-x}}<\varepsilon$. Then argue that ${\mathcal {C}}=\{[x,y]:x\in E,x<y<y_{x}\}$ is a Vitali cover of E. Let $\delta \in {\mathbb {R}}^{+}$ be the value corresponding to $\varepsilon$ in the definition of absolute continuity. Use the Vitali Covering Lemma to find disjoint $I_{1},\dots ,I_{n}\in {\mathcal {C}}$ such that $\lambda \left([a,b]\smallsetminus \bigsqcup _{k=1}^{n}I_{k}\right)<\delta$ Prove that the gaps between these intervals sum to less than $\delta$ and then apply absolute continuity to these. Adding together the deviations in f on $I_{1},\dots ,I_{n}$, and on the gaps, show that $|f(b)-f(a)|<\varepsilon +\varepsilon (b-a)$

Exercise 3. AC to FTC Let $f:[a,b]\to {\mathbb {R}}$. Show that if f is absolutely continuous then $\int _{(a,b)}f'=f(b)-f(a)$.

Exercise 4. AC iff Anti-derivative Let $f:[a,b]\to {\mathbb {R}}$. Show that f is absolutely continuous if and only if there is a function $g:[a,b]\to {\mathbb {R}}$ such that $f(x)=f(a)+\int _{(a,x)}g$ Hint: the "only if" part is essentially just the result of Exercise 3 where the end of the interval is now used as a variable to define a function. For the "if" part, you essentially just need the additivity of integration, and a result from earlier which showed that one can make integral small by making its domain small.