Proving the Lebesgue measure space completes the Borel measure space

I am trying to complete Exercise 1.4.27 page 78 of Terence Tao's book "An Introduction to Measure Theory", but is caught up in some troublesome details. The exercise is as follows:

$$\bullet$$ Show that the Lebesgue measure space $$(\mathbb{R}^d,\mathcal{L}[\mathbb{R}^d],m)$$ is the completion of the Borel measure space $$(\mathbb{R}^d,\mathcal{B}[\mathbb{R}^d],m)$$

I know that the completion of a measure space $$(X,\mathcal{B},\mu)$$ is the measure space $$(X,\overline{\mathcal{B}},\overline{\mu})$$ where $$\overline{\mathcal{B}}=\{B\cup N |B \in \mathcal{B} \land N \in \mathcal{N} \}$$, where $$\mathcal{N}$$ is the collection of subsets of the null sets of $$\mathcal{B}$$, and $$\overline{\mu}:B\cup N \mapsto \mu(B)$$.

Claim: $$\mathcal{L}[\mathbb{R}^d] \subseteq \{B \cup N | B \in \mathcal{B}[\mathbb{R}^d] \land N \in \mathcal{N}\}$$, where $$\mathcal{N}:$$ Borel subnull-sets

Proof attempt:

Part 1: We show that any $$S \in \mathcal{L}[\mathbb{R}^d]$$ is of the form $$B \cup N$$, where $$B$$ is a Borel set and $$N$$ is a (Lebsegue) null set.

Let $$S \in \mathcal{L}[\mathbb{R}^d]$$. Suppose that $$m(S)=\delta > 0$$. Then there is a closed set $$E \subseteq S$$ such that $$m(S \setminus E)\leq \delta$$. Since any closed set is a Borel-set, this proves that $$\Lambda=\{U\subseteq S| U \in \mathcal{B}[\mathbb{R}^d]\}$$ is nonempty. Therefore, let $$B=\bigcup_{A \in \Lambda} A$$ be the largest Borel set contained in $$S$$. Then we also have $$B \in \mathcal{L}[\mathbb{R^d}]$$, and accordingly $$S \setminus B \in \mathcal{L}[\mathbb{R^d}]$$. We claim that $$m(S \setminus B)=0$$. Suppose not, and that $$m(S \setminus B)=\gamma >0$$. Then we can once again find a closed set $$F \subseteq S \setminus B$$, contradictiong the maximality of $$B$$. Therefore $$m(S\setminus B)=0.$$

Part 2: We show that any (Lebesgue) null set is a Borel subnull set.

Let $$N$$ be any (Lebsegue) null set. Then, given any $$n \in \mathbb{N}$$, there is an open set $$U_n$$ containing $$N$$, such that $$m(U_n \setminus N)\leq 2^{-n}$$. Let $$U=\cap_{n=1}^{\infty}U_n$$. Then $$U$$ is an open set containing $$N$$, furtheremore $$m(U)\leq 2^{-n}$$ for any $$n$$, so $$m(U)=0$$. Thus, $$U$$ is a (Borel) null set, showing that $$N \in \mathcal{N}$$.

The problem however, is in part 1 of my attempted proof, as $$B$$ is defined as an (possibly uncountable) union of Borel sets, so I cannot guarantee that it is even well defined. I tried to change the definition of $$B$$ to the union of all open sets contained in $$S$$, i.e. $$B:=S^\circ$$, but this makes matters worse, as $$m(S)>0$$ does not imply that $$S^\circ \neq \emptyset$$. Can anyone see a way around this problem?

• Is this useful for you? Dec 10, 2019 at 9:51
• No, unfortunately not. Dec 11, 2019 at 9:39

By Exercise 1.4.26 the completion of $$\mathcal{B}(\mathbb{R}^d)$$ is given by $$\overline{\mathcal{B}}(\mathbb{R}^d)= \left\{B\cup N: B\in\mathcal{B}(\mathbb{R}^d), N\in\mathcal{\overline{N}} \right\}$$ where $$\mathcal{\overline{N}}$$ is a collection of sub-null sets of the Borel measure space. We demonstrate that $$\overline{\mathcal{B}}(\mathbb{R}^d)=\mathcal{L}(\mathbb{R}^d)$$.

• $$\subseteq$$) Let $$\overline{E}\in\overline{\mathcal{B}}$$, i.e., $$E = B \cup N$$ for a Borel measurable $$B\in\mathcal{B}$$ and a Borel sub-null set $$N\subseteq N'\in\mathcal{N}$$. The former is contained in $$\mathcal{L}(\mathbb{R}^d)$$ since every Borel measurable set is Lebesgue measurable (p. 72). The latter is a Lebesgue null set, since by the monotonicity of the Lebesgue outer measure we have $$m^*(N)\leq m(N) = 0$$. Thus, their union is a Lebesgue measurable set.
• $$\supseteq$$) Let $$E \in \mathcal{L}(\mathbb{R}^d)$$. By Exercise 1.2.19 $$E$$ is a $$G_\delta$$ set with a Lebesgue null set removed, i.e., $$E=\bigcup_{n=1}^{\infty}U_n \ N$$ for open sets $$(U_n)_{n\in\mathbb{N}}$$ and a Lebesgue null set $$N$$. The former intersection is a Borel measurable set, and is thus contained in $$\mathcal{B}(\mathbb{R}^d)$$. The latter is a Borel sub-null set. This is because it is a Lebesgue null set, i.e., by definiton for each $$n\in\mathbb{N}$$ we can find an open set $$V_n$$ such that $$N\subseteq V_n$$ and $$m(V_n) = m(V_n/N)\leq 1/n$$. Then, $$\bigcup_{n=1}^{\infty}V_n$$ is a Borel measurable set, and it is in particular a Borel null set. But we have $$N\subseteq \bigcup_{n=1}^{\infty}V_n$$ and thus $$N$$ is a Borel sub-null set. Thus the union of both sets is contained in $$\overline{B}$$.

proof:

By the previous exercise we know that $$\overline{\mathcal{B}[\mathbb{R}^d]} = \{A \subset \mathbb{R}^d: \exists B, C \in \mathcal{B}[\mathbb{R}^d], A \triangle B \subset C \text{ with } m(C) = 0\}$$. We want to show that this space is exactly $$\mathcal{L}[\mathbb{R}^d]$$.

We already know that $$\mathcal{B}[\mathbb{R}^d] \subset \mathcal{L}[\mathbb{R}^d]$$. Let $$A \in \overline{\mathcal{B}[\mathbb{R}^d}]$$ and $$B, C \in \mathcal{B}[\mathbb{R}^d]$$ be as in the definition. By the fact that $$A \triangle B \subset C$$ we have $$B \setminus C \subset A \subset B \bigcup C$$, from which follows $$A = (B \setminus C) \bigcup (C \cap A)$$, which is Lebesgue measurable. So $$\overline{\mathcal{B}[\mathbb{R}^d}] \subset \mathcal{L}[\mathbb{R}^d]$$.

Conversely, Let $$E \subset \mathbb{R}^d$$ be a Lebesgue measurable subset. By Exercise $$1.2.19$$, $$E = E' \setminus N$$, where $$E'$$ is a $$G_\sigma$$ set (hence a Borel set)and $$N$$ a Lebesgue null set. Suppose that $$E' = \bigcap\limits^{\infty}_{n=1}U_n$$, $$U_n$$'s are open. Then we have: $$E \triangle E' = E' \setminus E = E' \setminus (E' \setminus E) = E' \bigcap N = (\bigcap\limits^{\infty}_{n=1}U_n) \bigcap N = \bigcap\limits^{\infty}_{n=1}(U_n \bigcap N)$$. Now, since $$U_n \bigcap N$$ is Lebesgue measurable for all $$n$$, by outer approximation by open set(Exercise $$1.2.7$$) we can have: For all $$n \geq 1$$, $$U_n \bigcap N \subset V_n$$, $$V_n$$ is open, and $$m(V_n \setminus U_n \bigcap N) \leq 1/n$$. Then $$\bigcap\limits^{\infty}_{n=1}(U_n \bigcap N) \subset \bigcap\limits^{\infty}_{n=1}V_n = V$$, which is a Borel set since it's an intersection of open sets. Note that $$m(V \setminus \bigcup\limits^{\infty}_{n=1}U_n \bigcap N) \leq 1/n$$ for all $$n$$ and thus $$m(V \setminus \bigcup\limits^{\infty}_{n=1}U_n \bigcap N) = 0$$. Together with the fact that $$\bigcup\limits^{\infty}_{n=1}(U_n \bigcap N)$$ is a null set, we see that $$V$$ is a Borel null set by monotonicity. We then conclude that $$E \in \overline{\mathcal{B}[\mathbb{R}^d}]$$ and $$\mathcal{L}[\mathbb{R}^d] \subset \overline{\mathcal{B}[\mathbb{R}^d}]$$.

In other words, $$\overline{\mathcal{B}[\mathbb{R}^d}] = \mathcal{L}[\mathbb{R}^d]$$.

• The fact that $C \bigcap A$ is Lebesgue measurable comes form the fact that any set of Lebesgue outer measure 0 is measurable. See (iii) of Lemma $1.2.13$. Feb 9 at 19:06