The Completion of a Measure Space.

Question

Let $(X, \mathcal{T}, \mu)$ be a measure space and $$\mathcal{N} = \{N\subset X: \exists A\in \mathcal{T}\ \textrm{such that}\ N\subset A\wedge \mu(A)=0\}.$$ Let's define $$\mathcal{T}_1 = \{B\cup N: B\in \mathcal{T}\wedge N\in\mathcal{N}\}.$$ Prove that $\mathcal{T}_1=\dot{\mathcal{T}}=\sigma(\mathcal{T}\cup \mathcal{N})$.

Can someone help me out with this?

Answer 1

Hints:

1. Prove that $\mathcal T_1$ is a $\sigma$-algebra. [This is the crucial part.]
2. Obviously, $\mathcal N\cup\mathcal T\subseteq\mathcal T_1$.
3. Whenever $\mathcal N\cup\mathcal T\subseteq\mathcal A$ for a $\sigma$-algebra $\mathcal A$, we have $\mathcal T_1\subseteq\mathcal A$.