# Extension of nullset with the Dirac measure

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 $$\mathbb{N}_{\delta_a}=\left\{ N \subseteq \mathcal{X} \hspace{0.1cm}|\hspace{0.1cm} \exists B \in \mathbb{B}\hspace{0,1cm} \text{such that} \hspace{0,1cm} N\subseteq B \hspace{0,1cm} \text{and} \hspace{0,1cm} \delta_a(B)=0 \right\}$$

(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 $$\mathcal{P}(X) \hspace{0.1cm} \text{is all subsets of} \hspace{0.1cm}\Bbb R$$.

$$\mathbf{Attempt}$$

(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.

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.