Measure space and nullsets

Sep 2016
16
0
Denmark
Hey,

I have the measure space \(\displaystyle (\mathcal{X},\mathbb{E},\mu)\).

I am considering the paving \(\displaystyle \mathbb{N}_{\mu}\) of all nullsets given by: \(\displaystyle \mathbb{N}_{\mu}= \left\{ N \in \mathcal{X} : \exists N \subseteq E \hspace{0,1cm} \text{and} \hspace{0,1cm} \mu(E)=0 \right\}\)

I am also considering $\mathbb{E}_{\mu}$ which is an extension of \(\displaystyle \mathbb{E}\) by the nullsets: \(\displaystyle \mathbb{E}_{\mu}= \left\{ E \cup N : E \in \mathbb{E} \hspace{0,1cm} \text{and} \hspace{0,1cm} N \in \mathbb{N}_{\mu} \right\}\)

A paving is understood to be an arbitrary collection of subsets.

I wish to show that a) \(\displaystyle \mathbb{E} \subseteq \mathbb{N}_{\mu}\) and b) \(\displaystyle \mathbb{N}_{\mu} \subseteq \mathbb{E}_{\mu}\).

Here is my attempt:

a)

If \(\displaystyle E \in \mathbb{E}\) then \(\displaystyle E \in \mathbb{E}_{\mu}\) since \(\displaystyle \mathbb{E}_{\mu}\) is an extension of \(\displaystyle \mathbb{E}\).

Or I could say:

Assume that \(\displaystyle E \in \mathbb{E}\). Then \(\displaystyle E \in \mathbb{E}_{\mu}\) since \(\displaystyle \mathbb{E}_{\mu}\) consists of elements \(\displaystyle E \cup N\) where \(\displaystyle E \in \mathbb{E}\).

b)

Assume that \(\displaystyle N \in \mathbb{N}_{\mu}\). Then \(\displaystyle N \in \mathbb{E}_{\mu}\) since \(\displaystyle \mathbb{E}_{\mu}\) consists of elements \(\displaystyle E \cup N\) where \(\displaystyle N \in \mathbb{N}_{\mu}\).

Both explanations a) and b) seem very simple and trivial (the "then" does not satisfy me)! But is the reasoning correct?

Appreciate corrections and guidance.