Let $\lambda$ be the Lebesgue measure on $[0, 1]$ and $\mathcal{A}$ be the $\sigma$-algebra of Lebesgue measurable sets in $[0, 1]$. Suppose that a set $N \subset [0, 1]$ is not in $\mathcal{A}$. Prove that $\lambda$ can be extended to a measure on a $\sigma$-algebra $\mathcal{B}$ that contains $\mathcal{A}$ and $N$.
Is the below $\textbf{attempt}$ correct:
I only need to show that $\lambda$ is measure on the $\sigma$-algebra $\mathcal{B}=\sigma\big(\mathcal{A} \cup\{N\}\big)$.
let $E\in \mathcal{B}$, given that $\lambda$ is measure on $\mathcal{A}$ it follows that $\lambda\geq0$ and $\lambda(\phi)=0$ is satisfied. So I only need to show that it also satisfies the $\sigma$-additivity. I remember the following lemma from of my homeworks;,
$\textbf{BUT}$ have no intuition about the lemma, (how would I can think of solving this question without knowing the following lemma?)
lemma: All sets $E \in \sigma\big(\mathcal{A} \cup\{N\}\big)$ can be written as $E = (A\cap N) \cup (B\cap N^c)$ where $A,B \in \mathcal{A}$. (I skip the proof, but can be found here)
So given above; let $\{E_i\}_{i\in \mathbb{N}}$ be a collection of disjoint sets in $\mathcal{B}$.
\begin{align} \lambda \bigg(\bigcup_{i\in \mathbb{N}}E_i\bigg) & = \lambda \bigg(\bigcup_{i\in \mathbb{N}}A_i\cap N) \cup (B_i\cap N^c)\bigg)\\ & = \sum_{i\in \mathbb{N}}\lambda \bigg((A_i\cap N) \cup (B_i\cap N^c)\bigg)\\ & = \sum_{i\in \mathbb{N}}\lambda (E_i) \\ \end{align}
where the second equality follows from the fact $(A_i\cap N),(B_i\cap N^c)\in \mathcal{A}$, as $A_i,B_i \in \mathcal{A}$ and $\lambda$ is measure on $\mathcal{A}$. Also that $(A_i\cap N) \cap (B_i\cap N^c)= \phi $