# The restriction of Lebesgue measure to the $\sigma$-algebra of Borel subsets of $\mathbb{R}$ is not complete.

I am new to measure theory. I am confused by the following claim from Measure Theory by Donald Cohn (Section 1.5 Completeness and Regularity):

Claim$$\quad$$ The restriction of Lebesgue measure to the $$\sigma$$-algebra of Borel subsets of $$\mathbb{R}$$ is not complete.

In this post, I will denote the Lebesgue outer measure by $$\lambda^*$$.

According to the textbook,

Definition$$\quad$$ The restriction of Lebesgue outer measure on $$\mathbb{R}$$ (or on $$\mathbb{R}^d$$) to $$\mathcal{B}(\mathbb{R})$$ or to $$\mathcal{B}(\mathbb{R}^d)$$ is called Lebesgue measure and will be denoted by $$\lambda$$.

The restriction of Lebesgue outer measure on $$\mathbb{R}$$ (or on $$\mathbb{R}^d$$) to the collection of Lebesgue measurable subsets of $$\mathbb{R}$$ (or of $$\mathbb{R}^d$$) is also called Lebesgue measure and will be denoted by $$\lambda$$ as well.

Definition$$\quad$$ Let $$(X,\mathcal{A},\mu)$$ be a measure space. The measure $$\mu$$ (or the measure space $$(X,\mathcal{A},\mu)$$) is complete if the relations $$A\in\mathcal{A}$$, $$\mu(A)=0$$, and $$B \subseteq A$$ together imply that $$B\in\mathcal{A}$$.

So, my understanding of this claim is the following:

Let $$(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$$ be a measure space. Let $$A$$ be a Borel subset of $$\mathbb{R}$$ with $$\lambda(A)=0$$. There exists a subset $$B$$ of $$A$$ such that $$B \notin \mathcal{B}(\mathbb{R})$$.

I saw that this post used a Vitali set as an example to show that such a set $$B$$ does exist. However, I am very confused by its explanation. Let $$V$$ be a Vitali set in $$[0,1]$$. I know that, since $$V$$ is not measurable with respect to the Lebesgue outer measure $$\lambda^*$$, it follows that $$V \notin \mathcal{B}(\mathbb{R})$$. I think we would want to find a set $$A$$ such that $$A \in \mathcal{B}(\mathbb{R})$$, $$\lambda(A)=0$$, and $$V \subseteq A$$. But what exactly is this $$A$$ in the example?

In that post I mentioned above, it seems to me that it wanted to say that the Vitali set $$V$$ is of Lebesgue measure zero in $$\mathbb{R}^2$$ and itself is not a Borel subset of $$\mathbb{R}^2$$. However, if $$V$$ is not even Lebesgue measurable (i.e., $$V$$ is not measurable with respect to the Lebesgue outer measure), wouldn't $$\lambda(V)$$ be not defined? In addition, I couldn't see how his example showed the claim is true, because there was no such set $$A \subseteq \mathcal{B}(\mathbb{R})$$.

Basically, I am completely lost. I would really appreciate it if someone could help me clarify it or present another example!

As @LeeMosher pointed out. The claim should have been understood in the following way:

Let $$(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$$ be a measure space. For some Borel subset $$A$$ of $$\mathbb{R}$$ with $$\lambda(A) = 0$$, there exists a subset $$B$$ of $$A$$ where $$B \notin \mathcal{B}(\mathbb{R})$$.

• In what you write regarding your understanding of this claim, by writing Let $A$ be... you are implicitly using a universal quantifier on $A$, which is wrong. The correct statement of the claim is that there exists $A$, and there exists $B$, such that $A \in \mathcal A$, and $\lambda(A)=0$ and $B \subset A$ and $B \not\in \mathcal A$. This statement becomes false when there exists $A$ is replaced by for all $A$. For a counterexample, take any $A \in \mathcal B(\mathbb R)$ which is countable: $\lambda(A)=0$, and every one of its subsets is countable and hence a Borel set. Commented Mar 30 at 19:34
• @LeeMosher Thank you very much for pointing it out! Commented Mar 30 at 22:37

The answer in question constructs a set, namely $$B := V \times \{0\}$$ where $$V$$ is a (one dimensional) vitali set in $$[0,1]$$, which is not borel measurable ($$B\notin\mathcal{B}(\mathbb{R}^2)$$).
It is clear that $$B\subseteq [0,1]\times \{0\}=:A$$ and since $$A$$ is clearly a zero set with regards to the 2-dimensional Lebesgue measure, $$B$$ is also Lebesgue measurable as a subset of a Lebesgue measurable zero set. So we have $$A\in \mathcal{B}(\mathbb{R}^2)$$, $$\lambda(A)=0$$ and $$B\subseteq A$$ but $$B\notin \mathcal{B}(\mathbb{R}^2)$$.
To show that $$B$$ is not borel measurable ($$B\notin \mathcal{B}(\mathbb{R}^2)$$), the answer makes use of the map $$f: \begin{cases} \mathbb{R}\to \mathbb{R}^2\\ x \to (x,0) \end{cases}$$ which is continuous and therefore Borel measureable. Then $$V=f^{-1}(B)$$ so $$V$$ would be measurable if $$B$$ were measurable, which is a contradiction so $$B$$ can not be Borel measurable.
Let $$(\mathbb{R}^2,\mathcal{B}(\mathbb{R}^2),\lambda)$$, be a measure space. Let $$A$$ be a Borel subset of $$\mathbb{R}^2$$ with $$\lambda(A)=0$$. There exists a subset $$B$$ of $$A$$ such that $$B\notin\mathcal{B}(\mathbb{R}^2).$$