Lebesgue's Theorem

Lebesgue showed that a striking result that a monotone function is differentiable almost everywhere on an interval. We begin the discussion from a key concept, Vitali Cover.

Definition Vitali Cover

1. $F:$ a collection of closed, bounded non-degenerate(of positive length) intervals.
2. $E:$ a set.
3. Then $F$ is a Vitali cover of $E$ if $\forall x \in E$, $\forall \epsilon > 0$, $\exists$ interval $I \in F$, s.t., $x \in I$ and $\mu(I) < \epsilon$.

Theorem Vitali Covering Lemma

1. $E$ a set of finite outer measure.

2. $F$ a Vitali cover of $E$.

3. $\forall \epsilon > 0$, $\exists$ a finite disjoint collection of intervals $\{ I_k \} _{k = 1}^n$ of $F$ for which

   $$\mu^*(E \setminus \cup_{k = 1}^n I_k) \le \epsilon$$

Proof: Since $E$ is of finite outer measure, there is a open set $O$ of finite measure that contains $E$ for which $\mu(O) \le \mu^*(E) + \epsilon$. For each $x \in E$, we have $x \in O$. As $O$ is open, $\exists \epsilon > 0$, s.t., $N_\epsilon(x) \subset O$. Because $F$ is a Vitali cover of $E$, there exists an interval $I$ that contains $x$ with length $\mu(I) < \epsilon / 2$. This means $I \subset N_\epsilon(x) \subset O$. Therefore, we may assume that $F$ contains only intervals within $O$.

Then we are going to select a disjoint countable (either finite or infinite) sub-collection of intervals from $F$. Suppose $n$ is a natural number and the collection $\{ I_k \} _{k = 1 }^n$ of $F$ has been chosen. If $\{ I_k \} _{k = 1}^n$ covers $E$, then we are done. Otherwise, we pick a new interval $I_{k + 1}$ from $F$ according to two criteria:

1. $I_{k + 1}$ should be disjoint with the collection $\{ I_k \}_{k = 1}^n$, i.e., $I_{k + 1} \cap (\cup_{k = 1}^n I_k) = \emptyset$.
2. The length of $I_{k + 1}$ should be as large as possible.

Such $I_{k + 1}$ always exists. Let $x \in E \setminus \cup_{k = 1}^n I_k$. As $\cup_{k = 1}^n I_k$ is closed, $O \setminus \cup_{k = 1}^n I_k$ is open. Therefore, there is an interval in $F$ that is inside that open set and contains $x$. We can now define the nonempty collection of such intervals in $F$ as $F_n$:

As all interval in $F_n$ are contained in $O$ with finite measure,

exists and is a real number. Although we may not be able to find an interval $I$ that achieves the maximum length, i.e., $\mu(I) = \alpha_n$, it suffices to find one that is "a little bit" shorter. The criteria is not unique and for our proof we choose an $I_{k + 1}$ of measure $\mu(I_{k + 1}) > \alpha_n / 2$.

Now $\{ I_k \}_{k = 1}^\infty$ is a disjoint collection whose union is within $O$. By the countable additivity and monotonicity of measure,

We infer that $\mu(I_k) \rightarrow 0$. Moreover, $\{ I_k \}_{k = 1}^\infty$ has the following property: for each interval $I \in F$ that contains an element $x \in E$, either it belongs to or overlaps with $\{ I_k \}_{k = 1}^\infty$. Otherwise, $I$ is disjoint with $\{ I_k \}_{k = 1}^\infty$. But the selection process guarantees that $\mu(I) \le 2 \mu(I_k)$ for all $k$, which implies $\mu(I) = 0$, a contradiction (any $I \in F$ must have positive length).

We claim that for any natural number $n$, it holds that

As before, for $x \in E \setminus \cup_{k = 1}^n I_k$, $\exists \ I \in F_n$ for which $x \in I$. Then $I$ must belongs to or intersect with $\cup_{k = n + 1}^\infty I_k$. Let $m$ be the first natural number that $I_{m} \cap I \neq \emptyset$. By the selection process of $I_m$, $\mu(I_m) \ge \mu(I) / 2$. This means $I \subset 5 * I_m$.

Let $\epsilon > 0$, $\exists N > 0$, s.t., $\sum_{k = N}^\infty \mu(I_k) < \epsilon / 5$. This choice of $N$, together with the fact that $E \setminus \cup_{k = 1}^n I_k \subset (\cup_{k = n + 1}^\infty 5 * I_k)$, establishes the desired result.

$\blacksquare$

Lemma

Let $f : [a, b] \rightarrow R$ be a non-decreasing function, then for each $\alpha > 0$,

and

Proof:
Define $E = \{ x \in [a, b] : \bar D f(x) \ge \alpha \}$. For each $0 < \lambda < \alpha$, let $F_{\lambda} = \{ [c, d] \subset [a, b], f(d) - f(c) \ge \lambda (d - c) \}$. Since $\bar D f \ge \alpha$ on $E$, $F$ is a Vitali cover of $E$. By Vitali Covering Theorem, there is a finite disjoint sub-collection of intervals $\{ [c_k, d_k] \}_{k = 1}^n$ of $F$ for which

Therefore,

Taking the limit of $\lambda \rightarrow \alpha$, and $\epsilon \rightarrow 0$, we proves

$\blacksquare$.

Theorem Lebesgue

If a function $f$ is monotone on a interval $(a, b)$, then it is differentiable almost everywhere.

Proof:
Without lost of generality, assume $f$ is increasing. The set of non-differentiable points can be written as

where $Q$ is the set of rational numbers. It suffices to consider just $E_{\alpha, \beta} = \{ x \in (a, b) : \bar D f(x) \ge \alpha > \beta > \underline D f(x) \}$. We prove that $E_{\alpha, \beta}$ has zero measure and therefore by countable additivity $E$ has measure zero.

For fix $E_{\alpha, \beta}$, choose an open set $O$ for which $E_{\alpha, \beta} \subset O \subset (a, b)$ and $\mu^*(O \setminus E_{\alpha, \beta}) \le \epsilon$. Let $F_{\beta} = \{ [c, d] \subset O, f(d) - f(c) \le \beta (d - c) \}$ be a Vitali cover of $E_{\alpha, \beta}$. By Vitali Covering Theorem, there is a finite disjoint sub-collection of intervals $\{ [c_k, d_k] \}_{k = 1}^n$ for which

Therefore,

On the other hand, for each interval $[c_k, d_k]$, by the previous lemma,

Together with sub-additivity of outer measure,

Combined, we obtained

By arbitrariness of $\epsilon$, it holds that

It concludes that $\mu^*(E_{\alpha, \beta}) = 0$.

$\blacksquare$.

Reference

1. H.L.Royden, P.M. Fitzpatrick, Real Analysis