# $\sigma$ algebra and sets of measure $0$

1. A collection of measurable sets follows a $\sigma$ algebra.

2. All sets of measure $0$ is measurable.

My questions are:

1. What's the significance of sets of measure $0$ in the $\sigma$ algebra of measurable sets? Are they some sort of "identity"?

2. What's the significance of the proposition:"the translate of a measurable set is measurable." in the set of measurable set's $\sigma$ algebra? Does it somehow suggest an equivelence relation?

3. Does the previous proposition and sets of measure $0$ carry any interesting effect?

• Do you with "All sets of measure $0$ is measurable" refer to that if $E \subset F$ and $F$ is measurable with $m(F)=0$ (where $m$ is Lebesgue measure) then $E$ is measurable (of course with $m(E)=0$)? – md2perpe Sep 8 '18 at 18:33
• @md2perpe Real Analyssi fourth edition by Royden and Fitzpatrick section 2.4 Theorem 11's proof the last paragraph. All sets of measure $0$ is measurable(Lebesgue), and it's not hard to prove. – user416486 Sep 8 '18 at 19:41
• Ah, it refers to that a set $N$ of outer measure $0$ is measurable, i.e. that $m^*(A) = m^*(E \cap N) + m^*(E \cap N^c)$ for all sets $A$. – md2perpe Sep 8 '18 at 20:03

1. Sets of measure $0$, called null-sets, are primarily significant in that functions that only differ on a null-set work exactly the same under integration. This is an equivalence relation. Such functions will later be identified. You will also see a lot of statements saying that something is valid almost everywhere meaning that it's valid except perhaps on a null-set.