# Help save this proof about the regularity of the lebesgue measure on $\mathbb{R}^d$

In a previous homework our task was to prove the regularity of the Lebegue-measure on $$\mathbb{R}^d$$. More precisely:

Let $$(\mathbb{R}^d, \mathcal{M}^*, \lambda)$$ be a measure space and $$\mathcal{M}^* := \mathcal{M}^*(\mathbb{R}^d)$$ be the $$\sigma$$-algebra of the $$\lambda^*$$-measurable sets, with $$\lambda$$ being the by the Lebesgue-outer-measure $$\lambda^*$$ induced measure.

Show that \begin{align*} \lambda(A) & = \inf\{\lambda(O) \mid O \subset \mathbb{R}^d \text{ is open and } A \subset O \} \\ & = \sup\{\lambda(K) \mid K \subset \mathbb{R}^d \text{ is compact and } K \subset A \}. \end{align*}

Our attempt goes as follows:

For all $$A \in \mathcal{M}^*$$ we want to show

$$\begin{equation*} \lambda(A) \ge \inf\{ \lambda(O): O \subset \mathbb{R}^d \text{ open und } A \subset O \} \ge \sup\{ \lambda(K): K \subset \mathbb{R}^d \text{ compact and } K \subset A \} \ge \lambda(A) \end{equation*}$$

First inequality: If $$\lambda(A) = \infty$$, because of the monotonicity of the measure, we have $$\lambda(O) = \infty$$ for all open sets $$O$$ with $$A \subset O$$.

If $$\lambda(A) < \infty$$, we define $$A_{\delta} := \bigcup_{a \in A} U_{\delta}(a)$$, which is a open and therefore measurable set with $$A \subset A_{\delta}$$.

Now we have $$A_{\delta} \setminus A \xrightarrow{\delta \to 0} \emptyset$$ and because of the continuity from below in the emptyset of $$\lambda$$, and, because $$A_{\delta} \setminus A$$ is measurable, because it's the difference of measurable sets, $$\begin{equation*} \forall \varepsilon > 0 \ \exists \delta > 0: \lambda(A_{\delta} \setminus A) < \varepsilon. \end{equation*}$$

Because $$A_{\delta}$$ is open and $$\sigma$$-additivity of $$\lambda$$ we have $$\begin{equation*} \inf\{ \lambda(O): O \subset \mathbb{R}^d \text{ open und } A \subset O \} \le \lambda(A_{\delta}) = \lambda(A_{\delta} \setminus A) + \lambda(A) < \varepsilon + \lambda(A). \end{equation*}$$

Because $$\varepsilon > 0$$ was arbitrary, the inequality follows.

The second inequality follows from the monotonicity of $$\lambda^*$$: From $$K \subset A \subset O$$ we have $$\lambda(K) \le \lambda(O)$$ for $$K$$ und $$O$$ as defined above. Because taking the supremum or infimum doesn't change weak inequalities, the inequality follows.

The third inequality: One can show, that every open set is $$\sigma$$-compact.

Lemma Let $$A_{\delta} \subset \mathbb{R}^d$$ be an open subset. Then there exists a countable family of closed bounded axially parallel cubes $$(W_k)_{k \in \mathbb{N}}$$, so that $$A_{\delta} = \bigcup_{n \in \mathbb{N}} W_n$$.

Proof Let $$a \in A_{\delta}$$. Because $$A_{\delta}$$ is open, there exists an $$\varepsilon_{a} > 0$$, so that $$U_{\varepsilon_a}(a) \subset A_{\delta}$$. For every $$a \in A_{\delta}$$ we choose a bounded closed axially parallel cube $$W_a$$ with rational midpoint from $$\mathbb{Q}^d$$ and rational edge length $$q \in \mathbb{Q}$$, so that $$\begin{equation*} a \in W_{a} \subset U_{\varepsilon_a}(a) \subset A_{\delta} . \end{equation*}$$ This is always possible, because $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$. Therefore, we have $$A_{\delta} = \bigcup_{n \in \mathbb{N}} W_n$$. Since there are only countable many cubes of this form, the union is countable. $$\square$$

We let $$K_n := \bigcup_{j = 1}^{n} W_j$$, which is compact as union of compact sets. Then $$K_n \xrightarrow{n \to \infty} A_{\delta}$$.

Now we have $$A_{\delta} \setminus K_n \xrightarrow{n \to \infty} \emptyset$$ and with analogous argumentation as above $$\begin{equation*} \forall \varepsilon > 0 \ \exists N \in \mathbb{N}: \lambda(A_{\delta} \setminus K_n) < \varepsilon \ \forall n > N. \end{equation*}$$ Therefore follows for all $$\varepsilon > 0$$ \begin{align*} \lambda(A) \le \lambda(A_{\delta}) \le \lambda(A_{\delta}\setminus K_n) + \lambda(K_n) < \varepsilon + \sup\{ \lambda(K): K \subset \mathbb{R}^d \text{ compact und } K \subset A_{\delta} \}, \end{align*}

Because $$\varepsilon > 0$$ was arbitrary and $$\delta$$ can be arbitrarily small, the inequality follows.

My problem

I assume the proof of the second inequality is right. But: I not sure if the reasoning at the end of the proof of the last inequality is rigorous enough. Especially, if you take $$A := \mathbb{Q}$$ in the proof of the first inequality, then for every $$\delta > 0$$, we have $$A_{\delta} = \mathbb{R}$$ and therefore, we don't have $$A_{\delta} \setminus A \to \emptyset$$ or even $$\lambda(A_{\delta} \setminus A) \to 0$$.

Is there anyway to ''save'' this proof by fixing it and not changing the approach?

Correct Proof

First inequality Case 1: $$\lambda(A) = \infty$$. As above.

Case 2: $$\lambda(A) < \infty$$.

Utilising the Caratheodory construction of the Lebesgue measure, we know that $$\forall \varepsilon > 0 \ \exists (a_n,b_n] := \prod_{i=1}^d \left(a_{n}^{(i)},b_{n}^{(i)}\right]: A \subset \bigcup_{n \in \mathbb{N}} (a_n,b_n] \quad \text{and} \quad \sum_{k=1}^\infty \lambda \left((a_k,b_k]\right) < \lambda(A) + \frac{\varepsilon}{2},$$ where the $$d$$-dimensional cubes $$(a_n,b_n]$$ are pairwise disjoint.

Now let $$U := \bigcup_{n=1}^\infty (a_n, b_n+ t_n \varepsilon) \qquad \text{with} \qquad t_n := 2^{-n-2d-1} \max\{1,b_{n}^{(i)} -a_{n}^{(i)}\}^{-(d-1)}.$$ Now we have $$A \subset \bigcup_{n=1}^\infty (a_n,b_n] \subset U$$ and $$\lambda(U) < \lambda(A) + \varepsilon$$.

Therefore, we have $$\begin{equation*} \inf\{ \lambda(O): O \subset \mathbb{R}^d \text{ open und } A \subset O \} \le \lambda(U) < \lambda(A) + \varepsilon. \end{equation*}$$ Because $$\varepsilon > 0$$ was arbitrary, the inequality follows.

Second inequality Same as above

Third inequality

From a previous homework we know that $$\mathcal{B}(\mathbb{R}^d) \subset \mathcal{M}^*$$ and therefore, that $$\begin{equation*} \forall M \in \mathcal{M}^* \ \exists B \in \mathcal{B}(\mathbb{R}^d), N \in \mathcal{N}: M = B \cup N, \end{equation*}$$ where $$\mathcal{N}$$ is the set of borel-null-sets. Now define $$\begin{equation*} \mathcal{D} := \{ B \in \mathcal{B}(\mathbb{R}^d): \lambda(B) = \sup\{\lambda(K) \mid K \subset \mathbb{R}^d \text{ is compact and } K \subset B \} \} \end{equation*}$$ as the set of all inner regular sets in the borel $$\sigma$$-algebra, which, by construction is a subset of $$\mathcal{B}(\mathbb{R}^d)$$.

Now we want to show, that $$\mathcal{B}(\mathbb{R}^d) \subset \mathcal{D}$$ to conclude $$\mathcal{B}(\mathbb{R}^d) = \mathcal{D}$$.

1. From the above lemma we know, that for all open sets $$\mathcal{O} \subset \mathbb{R}^d$$ we have $$\mathcal{O} \in \mathcal{D}$$, since by the continuity of the Lebesgue-measure, for all $$\varepsilon > 0$$ we have $$\lambda(A) \le \lambda(\bigcup_{n \in \mathbb{N}} W_n) + \varepsilon$$.

2. Since the set of open sets is a $$\cap$$-stable generator of $$\mathcal{B}(\mathbb{R}^d) \subset \mathcal{P}(\mathbb{R})^d$$, we only need to show that $$\mathcal{D}$$ is a dynkin system. (German Wikipedia article on this line of argument)

• We have $$\mathbb{R}^d \in \mathcal{D}$$, because $$\lambda(\mathbb{R}^d) = \infty = \sup\{\lambda(K): K \subset \mathbb{R}^d \}$$.

• Let $$D \in \mathcal{D}$$. Then ??

• Let $$\{ A_n \}_{n \in \mathbb{N}} \subset \mathcal{D}$$ be a family of disjoint subsets. Then

The first property is called 'outer regularity' and the second 'inner regularity'. First of all, you can not take $$A_\delta$$ as defined above, as you have already noted. Also the proof of the second equality (i.e. your argument in the section third inequality) is false, because it can happen that $$K_n$$ is not a subset of $$A$$! (Take your example: Then $$A_\delta = \mathbb{R}$$ and $$K_n$$ will approximate the measure of $$\mathbb{R}$$.)

To prove the 'outer regularity' use the Caratheodory construction of the Lebesgue measure: If $$\lambda(A) < \infty$$, then we can find $$(a_n,b_n] = \prod_{i=1}^d (a_{n,i},b_{n,i}]$$ disjoint ($$d$$-dimensional) cubes with $$\sum_{k=1}^\infty \lambda ((a_n,b_n]) < \lambda(A) + \varepsilon/2$$ Thus, we can take $$U= \bigcup_{n=1}^\infty (a_n,b_n+ t_n \varepsilon)$$. (Note $$A \subset \bigcup_{n=1}^\infty (a_n,b_n] \subset U$$ and if $$t_n$$ is taken appropriate, e.g. $$t_n = 2^{-n-2d-1} \max\{1,b_{n,i} -a_{n,i}\}^{-(d-1)}$$ we get $$\lambda(U) < \lambda(A) + \varepsilon$$.)

The proof ot the 'inner regularity' is more complicated. First recall that any set $$\mathcal{M}^*$$ is the completion of the Borel-$$\sigma$$-algebra. Thus any $$A \in \mathcal{M}^*$$ can be written as $$A= B \cup M$$ with a Borel set $$B$$ and $$M \subset N$$ with a Borel-nullset $$N$$. Thus, we only have to prove the 'inner regularity' for Borel-sets. Define the system $$\mathcal{D} := \{ B \in \mathcal{B}(\mathbb{R}^d) : B \text{ is inner regular}\}.$$

1. Your argument shows that open sets are in $$\mathcal{D}$$.
2. Check that $$\mathcal{D}$$ is a Dynkin-system.
3. Since the set of all open sets is a $$\cap$$-stable generator of the Borel-$$\sigma$$-algebra we can conclude that $$\mathcal{D} = \mathcal{B}(\mathbb{R}^d)$$.

I additionally added a prove of 2: Note that $$\mathbb{R}^d$$ is an open set. If $$(A_n)_{\in \mathbb{N}} \subset \mathcal{D}$$ are disjoint, then we can take compact $$K_n \subset A_n$$ with $$\lambda(A_n) < \lambda(K_n) + \varepsilon 2^{-n}$$. Since a finite union of compact sets is compact, but in general not an infinite union, we have to truncate the sums as follows.

First case: Now if $$\lambda(\cup_{n=1}^\infty A_n) = \sum_{n=1}^\infty \lambda(A_n) = \infty$$, then for any $$K>0$$ there exists $$N$$ such that $$\sum_{n=1}^N \lambda(A_n) >K.$$ Thus $$\sum_{n=1}^N \lambda(K_n) > K-\varepsilon$$. So taking the compact set $$K = \cup_{k=1}^N K_n$$ shows that the supremum is $$>K-1$$. Since $$K>0$$ was arbitrary, we get that the supremum is $$\infty$$.

In the second case, we have for some $$N \in \mathbb{N}$$, because the series is convergent, that $$\lambda(\cup_{n=1}^\infty A_n) < \sum_{n=1}^N \lambda(A_n) + \varepsilon.$$ Thus $$\lambda(\cup_{n=1}^\infty A_n) < 2 \varepsilon + \lambda(\cup_{n=1}^N K_n)$$ and therefore we can take the compact set $$K = \cup_{n=1}^N K_n$$. This proves that $$\cup_{n=1}^\infty A_n \in \mathcal{D}$$.

Let $$A,B \in \mathcal{D}$$ with $$B \subset A$$. For the next argument we need that both $$B$$ and $$A$$ have finite measure. Of course, this is satisfied if the sets are bounded. Thus take $$A_i = A \cap \prod_{i=1}^d (n_i,n_i+1] \quad \text{and} \quad B_i = B \cap \prod_{i=1}^d (n_i,n_i+1]$$ with $$n_i \in \mathbb{Z}$$ instead of $$A$$ and $$B$$ to get bounded sets. (If $$A_i \setminus B_i \in \mathcal{D}$$. then also the union of this disjoint sets is in $$\mathcal{D}$$ by the previous argument. Moreover, show that also $$A_i,B_i \in \mathcal{D}$$.)

So we can assume that both $$B$$ and $$A$$ are bounded. Now take an open set $$U$$ with $$B \subset U$$ and $$\lambda(U \setminus B) < \varepsilon$$ (this is possible, because we already know that $$\lambda$$ is outer regular) and a compact set with $$K \subset A$$ and $$\lambda(A \setminus K) < \varepsilon$$. Define $$L = K \setminus U$$. Then $$L$$ is compact and $$\lambda( (A \setminus B) \setminus L) \le \lambda(U \setminus B) + \lambda(A \setminus K) < 2 \varepsilon.$$ Therefore $$A \setminus B \in \mathcal{D}$$.

• For the first inequality, we had the idea that we can instead choose $A_{\delta}$ in such similar fashion to the way it is possible to cover the rationals of a compact interval in $\mathbb{R}$ with countable many open subsets, whose measure is smaller than an arbitrary $\varepsilon > 0$. Would this work? Jan 13, 2019 at 13:48
• Yes, I have edited my answer. In the case of the rational numbers you can, in fact, take for example $U = \bigcup_{n=1}^\infty (q_n -2^{-n} \varepsilon, q_n +2^{-n} \varepsilon)$, where $(q_n)_n$ is an enumeration of $\mathbb{Q}$. But this argument only applies for countable sets. If $A=C$ is a (thin) Cantor set, you have uncountable many intervals in your union, but $C$ is a nullset. Jan 13, 2019 at 13:54
• I have added the construction of the $W_j$ with the proof of the $\sigma$-compactness of the open set $A_{\delta}$. Can you please explain why it is a problem, that $K_n \not\subset A$? Jan 13, 2019 at 15:05
• The lemma after the 'third inequality' shows that, if $U$ is open, there exists a countable family of closed bounded axially parallel cubes $(W_k)_{k \in \mathbb{N}}$ such that $A=\bigcup_{n \in \mathbb{N}} W_n$. Thus, by the continuity of measures, we have $\lambda(U) < \lambda(\bigcup_{n=1}^N W_n) + \varepsilon$ for some $N \in \mathbb{N}$. Note that $K:=\bigcup_{n=1}^N W_n$ is a compact set. Jan 13, 2019 at 20:40
• Yes, the rest of your corrected proof is error-free. Note that the characterization of the completion (in the sense of $A= B \cup M$ and $B$ is Borel measurable and there exists a Borel-nullset $N$ with $M \subset N$) is not just a consequence of $\mathcal{B}(\mathbb{R}^d) \subset \mathcal{M}^*$, but a theorem about the characterization of the completion of a measure space. Jan 13, 2019 at 21:23