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?

  • $\begingroup$ Is this useful for you? $\endgroup$
    – drhab
    Dec 10, 2019 at 9:51
  • $\begingroup$ No, unfortunately not. $\endgroup$
    – AfterMath
    Dec 11, 2019 at 9:39

2 Answers 2


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 $\overline{B}$.


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

  • $\begingroup$ 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 $1.2.13$. $\endgroup$
    – shark
    Feb 9 at 19:06

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