Let $\delta_a$ be the Dirac measure on $(\Bbb R,\Bbb B)$, such that $\delta_a(B)=1$ if $a \in B$ and $0$ otherwise, and let $\mathbb{N}_{\delta_a}=\left\{ N \subseteq \mathcal{X} \hspace{0.1cm}|\hspace{0.1cm} \exists B \in \mathbb{B}\hspace{0,1cm} \text{such that} \hspace{0,1cm} N\subseteq B \hspace{0,1cm} \text{and} \hspace{0,1cm} \delta_a(B)=0 \right\}$

I am asked to show:

(1) $\Bbb{N}_{\delta_a}=\left\{A \subseteq \Bbb{R} \hspace{0.1cm}| \hspace{0.1cm} A \subseteq \Bbb{R} \backslash \{a\}\right\}$

(2) $\Bbb{B}_{\delta_a} = \mathcal{P}(X)$, where $\mathcal{P}(X) \hspace{0.1cm} \text{is all subsets of} \hspace{0.1cm}\Bbb R$.


(1) I want to show the double inclusion $\Bbb{N}_{\delta_a} \subseteq\left\{A \subseteq \Bbb{R} \hspace{0.1cm}| \hspace{0.1cm} A \subseteq \Bbb{R} \backslash \{a\}\right\}$ and

$\{A \subseteq \Bbb{R} \hspace{0.1cm}| \hspace{0.1cm} A \subseteq \Bbb{R} \backslash \{a\}\} \subseteq \Bbb{N}_{\delta_a}$

Let $\delta_a(B)=0$ then $\{a\} \notin B$, which implies that for all $N \in \mathbb{N}_{\delta_a} \{a\} \notin N$. Now take any $A \subseteq \Bbb{R}\backslash \{a\}$. By the nature of $\Bbb{B}$ there is $B$ such that $A \subseteq B$ and for all $A \hspace{0.3cm}{a}\notin A$. Thus every $A \in \Bbb{R}\backslash \{a\}$ is in $\Bbb{N}_{\delta_a}$

On the other hand, all $N \in \Bbb{N}_{\delta_a}$ must be contained in $\Bbb{R}\backslash\{a\}$.

This shows $\Bbb{N}_{\delta_a}=\left\{A \subseteq \Bbb{R} \hspace{0.1cm}| \hspace{0.1cm} A \subseteq \Bbb{R} \backslash \{a\}\right\}$

(This does not feel quite rigorous)

(2) I'm not sure where to start. I know that the Borel $\sigma\text{-algebra}$ on $\Bbb R$ does not equal the powerset on $\Bbb {R}$ (from: Why is the Borel Algebra on R not equal the powerset?) And I fail to see, how this extension remedies that. Am I missing a theorem?

Any help checking (1) or providing a hint for (2) is much appreciated.


Your answer for 1) is OK.

Answer for 2): let $B$ be any subset of $\mathbb R$. Then $B=\{a\} \cup (B\setminus \{a\})$. By 1), $B\setminus \{a\}$ is in $\mathbb N _{\delta_a}$. Hence $B$ is the union of Borel set and a null set under $\delta_a$. Be definition of the completion it follows that $B$ belongs to the completed sigma algebra. Hence every set is measuarble in the completion.

  • $\begingroup$ Thank you very much! $\endgroup$ – Thomas More Dec 2 at 23:26

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