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

Question

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!

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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 $B \notin \mathcal{B}(\mathbb{R})$.

Answer 1

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.

In essence it shows the following statement in two dimensions, not in one!

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 $B\notin\mathcal{B}(\mathbb{R}^2).$