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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

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1 Answer 1

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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 $A \setminus B \in \mathcal{D}$.

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  • $\begingroup$ 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? $\endgroup$
    – Ramanujan
    Jan 13, 2019 at 13:48
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    $\begingroup$ 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. $\endgroup$
    – p4sch
    Jan 13, 2019 at 13:54
  • $\begingroup$ 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 $K_n \not\subset A$? $\endgroup$
    – Ramanujan
    Jan 13, 2019 at 15:05
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    $\begingroup$ 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. $\endgroup$
    – p4sch
    Jan 13, 2019 at 20:40
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    $\begingroup$ 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. $\endgroup$
    – p4sch
    Jan 13, 2019 at 21:23

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