lebesgue-measuremeasure-theory
A set $E\subseteq \mathbb{R}$ is called measurable if for all $A\subseteq \mathbb{R}$
$$m^*(A)=m^*(A\cap E)+m^*(A\cap E^c)$$The set of all measurable sets is called Lebesgue sigma algebra
Does not sigma algebra has properties? this is a way to "build" Lebesgue sigma algebra from "another way" by measurable sets?
Let me take a sidestep. Trying to understand $\sigma$ algebra through Vitali sets and Lebesgue measure, etc. is, in my opinion, the wrong approach. Let me offer a much, much simpler example.
A probability space is a measure space $(X, \sigma, \mu)$ where $\mu(X)=1$.
What's the simplest thing you can model with probability theory? Well, the flip of a fair coin. Lets write out all the measure theoretic details.
What are the possible outcomes? Well you can get a heads, $H$, or a tail, $T$. Measure theoretically, this is the set $X$. That is, $X=\{H,T\}$.
What are the possible events? Well first of all nothing can happen or something can happen. This is a given in ANY probability space (or measure space). Also, you can get a heads or you can get a tail. This is the $\sigma$ algebra. That is, $\sigma=\{\emptyset, X, \{H\}, \{T\}\}$.
What is the (probability) measure? Well what are the probailities? First, what is the probability that NOTHING happens? Well, $0$. That is, $\mu(\emptyset)=0$. What is the probability that SOMETHING happens? Well, $1$. That is, $\mu(X)=1$ (this is what makes something a probability space). What is the probability of a heads or a tail? Well $1 \over 2$. That is, $\mu(\{H\})=\mu(\{T\})=\frac12$.
This is just filling out all the measure theoretic details in a very simple situation.
Okay. In this situation EVERY subset of $X$ is measurable. So still, who gives a damn about $\sigma$ algebras? Why can't you just say "everything is measurable" (which is a perfectly fine $\sigma$ algebra for every set, including $\Bbb{R}$!!) and totally forget this business about $\sigma$ algebras? Well lets very slightly modify the previous example.
Lets say you find a coin, and you're not sure if it's fair or not. Lets talk about about the flip of a possibly unfair coin and fill in all the measure theoretic details.
Again, what are the possible outcomes? Again $X=\{H,T\}$.
Now here is where things are interesting. What is the $\sigma$ algebra for this coin flip (the events)? This is where things differ. The $\sigma$ algebra is just $\{\emptyset, X\}$. Remember, a $\sigma$ algebra is the DOMAIN of the measure $\mu$. Remember that we don't know if the coin is fair or not. So we don't know what the measure (probability) of a heads is! That is, $\{H\}$ is not a measurable set!
What is the measure? $\mu(\emptyset)=0$ and $\mu(X)=1$. That is, something will happen for sure and nothing won't happen. What a remarkably uninformational measure.
Here is a situation where the $\sigma$ algebras ARE different because they describe two different situations. A good way to think about $\sigma$ algebras is "information", especially with probability spaces.
Lets connect this back to Vitali sets and Lebesgue measure.
AFAIK, I can have a sigma-algebra that contains Vitali set. Everything can be a sigma-algebra, as long as it satisfies its 3 simple axioms. So if there is something you could add to clarify the quoted explanation, I'd be very grateful.
You are quite right. You CAN have a $\sigma$ algebra of $\Bbb{R}$ that contains every Vitali set. Just like you can have a $\sigma$ algebra of $\{H,T\}$ that contains $\{H\}$. But the point is that you might not be able to assign this a measure in a satisfactory way! Meaning, that if you have a list of properties that a you want Lebesgue measure wants to satisfy, a Vitali set necessarily can't satisfy them. So you "don't know" what measure to assign to it. It is not measurable with respect to a specified measure. You can create all sorts of measures where Vitali sets are measurable (for example one where the measure of everything is $0$).
Let me motivate the axioms of a $\sigma$ algebra in terms of probability.
The first axiom is that $\emptyset, X \in \sigma$. Well you ALWAYS know the probability of nothing happening ($0$) or something happening ($1$).
The second axiom is closed under complements. Let me offer a stupid example. Again, consider a coin flip, with $X=\{H, T\}$. Pretend I tell you that the $\sigma$ algebra for this flip is $\{\emptyset, X, \{H\}\}$. That is, I know the probability of NOTHING happening, of SOMETHING happening, and of a heads but I DON'T know the probability of a tails. You would rightly call me a moron. Because if you know the probability of a heads, you automatically know the probability of a tails! If you know the probability of something happening, you know the probability of it NOT happening (the complement)!
The last axiom is closed under countable unions. Let me give you another stupid example. Consider the roll of a die, or $X=\{1,2,3,4,5,6\}$. What if I were to tell you the $\sigma$ algebra for this is $\{\emptyset, X, \{1\}, \{2\}\}$. That is, I know the probability of rolling a $1$ or rolling a $2$, but I don't know the probability of rolling a $1$ or a $2$. Again, you would justifiably call me an idiot (I hope the reason is clear). What happens when the sets are not disjoint, and what happens with uncountable unions is a little messier but I hope you can try to think of some examples.
I hope this cleared things up.
We have the standard Lebesgue measure $\mu$, that is defined on all Borel sets, and obeys translation invariance and interval-consistency (my name for $\mu((a,b])=b-a$ for all relevant intervals; you call it uniform in the comments) . Indeed there are $\mathfrak{c}$ many Borel sets on which this $\mu$ is then defined.
Independently of that we can define null sets as:
$A \in \mathscr{N}$ iff for every $\varepsilon >0$ we can find at most countably many open intervals $(a_n, b_n)$ such that $A \subseteq \bigcup (a_n, b_n)$ and $\sum_n |b_n - a_n| < \varepsilon$ and check that such subsets are closed under subsets and countable unions (they form a $\sigma$-ideal). There are $2^\mathfrak{c}$ many null sets (e.g. as the standard Cantor set is one and so are all of its subsets.)
It turns out we can extend $\mu$ to a measure on more subsets, namely take the collection of all sets of the form $B \cup N$ where $B$ is Borel, and $N$ is a null set and just define $\mu'(B \cup N) = \mu(B)$. One can prove that this is well-defined (e.g. if a Borel set happens to be a null set, $\mu$ was $0$ on it anyway, justifying the name) and greatly extends the domain of $\mu$, while keeping translation invariance.
The existence of a Vitali set (if we assume AC) shows that we cannot extend $\mu'$ even further to $\mathscr{P}(\mathbb{R})$ while keeping translation invariance intact. $\mu'$ is called the completion of $\mu$ and it's also obtained when we apply the Carathéodory theorem on the Lebesgue measure on the half-open interval algebra; we get (IIRC) the same class of Lebesgue measurable subsets.
It depends on your application whether you want to work with only Borel sets or all Lebesgue measurable sets. Mostly in analysis the latter is done; taking the larger domain.
If we give up translation invariance and so "neutralise" the Vitali set examples (whose proof of non-measurability hinges on this translation invariance property) there are still obstacles: under CH ($\mathfrak{c}=\aleph_1$) there can be no finite measure on $(0,1]$ that measures all subsets and that gives all singletons measure $0$ (much weaker than being uniform), as was shown by Ulam. So you could never define such a measure on all subsets without additional set-theoretic assumptions. See also real valued measurable cardinals etc. It gets complicated.
Best Answer
Yes, a $ \sigma- $ algebra has properties ! Let
$ \mathcal{L}= \{E \subseteq \mathbb R: m^*(A)=m^*(A\cap E)+m^*(A\cap E^c) \quad \forall A \subseteq \mathbb R\}.$
$ \mathcal{L}$ has the follwing properties (try a proof):
$ \mathbb R \in \mathcal{L}$;
$E\in \mathcal{L}$ implies $\mathbb R \setminus E \in \mathcal{L}$;
if $(E_j)$ is a sequence in $ \mathcal{L}$, then $\bigcup E_j \in \mathcal{L}.$
This shows that $ \mathcal{L}$ is a $ \sigma- $ algebra