Simple problem about borel and measurable sets

[hellbike]

Show, that Y(x(B)) = xY(B) (Y is Lebesgue_measure ) for every borel set B and x>0.
Show that also for measurable sets.

I don't know how to prove anything for neither borelian or measurable sets, so I'm asking someone for doing this problem, so i can do other problems with borelian and measurable sets by my own.

 

[micromass]

The method is always thesame: consider

$$\{A\subseteq \mathbb{R}~\vert~Y(xA)=xY(A)\}$$

and show that this is a $$\sigma$$-algebra that contains the open intervals...

 

[hellbike]

and this is for both, borel and measurable?


Is showing that this is true for approximation using open intervals is enough for proving this for borel sets?

because borel sets are these that can be approximated using open intervals, right?

And what are measurable sets?

 

[losiu99]

Intervals are enough for Borel sets because Borel $$\sigma$$-algebra is by definition the smallest one containing all the intervals. You still need to prove that this set is $$\sigma$$-algebra.

Borel sets are enough for measurable sets, because it's a completion of measure defined on Borel sets. Each measurable set is a union of a Borel set and set $$F$$ of (outer) measure 0. Set of outer measure 0 is contained in Borel set $$E$$ of measure 0, and so by the previous part, $$xE$$ is of measure 0. Hence, $$xF$$ also has outer measure 0.

 

[hellbike]

I'm not sure if i understand this, because this seems too easy.

It's obvious that Y(x(B)) = xY(B) is true for every interval.
It's obvious that this set is closed under countable sums from definition of measure (this just requires construction of pair disjoint sets).
A\B for every A,B from R is also obvious from Y(A\BnA) = Y(A) - Y(BnA)

And that's it?

 

[losiu99]

It seems fine to me.