Can a G-delta Set and a Borel Set Have the Same Lebesgue Measure?

[OMM!]

Homework Statement

Show there's a G-delta set B, with E [tex]\subseteq[/tex] B s.t.

[tex]\lambda[/tex](E) = [tex]\lambda[/tex](B)

Where [tex]\lambda[/tex] is the Lebesgue measure and E is a Borel set.

Homework Equations

- G-delta set is a countable intersection of open set.
- Lebesgue measure has properties: monotonicity, countable additivity, translation invariance, measure of countable subset is 0, correct length for closed intervals.

The Attempt at a Solution

I get the impression that by the Monotonicity of the Lebesgue measure, we can show [tex]\lambda[/tex](E) [tex]\leq[/tex] [tex]\lambda[/tex](B) as E [tex]\subseteq[/tex] B.

Now it is just a case of showing the reverse holds as well, to show the equality.

However, I have no idea where to start with this reverse inequality, or if this is even the correct approach to take. Helps!

 

[mighty2000]

Dear student,

Your approach is on the right track. To show the reverse inequality, we can use the countable additivity property of the Lebesgue measure. Let B = \bigcap_{n=1}^{\infty} U_n, where U_n are open sets. Since E \subseteq B, we can write E = \bigcap_{n=1}^{\infty} (E \cap U_n). Now, using the countable additivity property, we have:

\lambda(E) = \lambda\left(\bigcap_{n=1}^{\infty} (E \cap U_n)\right) = \lim_{N \to \infty} \lambda\left(\bigcap_{n=1}^{N} (E \cap U_n)\right)

Since each (E \cap U_n) is a Borel set, we can apply the Monotonicity property to get:

\lambda(E) \leq \lim_{N \to \infty} \lambda\left(\bigcap_{n=1}^{N} U_n\right) = \lambda\left(\bigcap_{n=1}^{\infty} U_n\right) = \lambda(B)

Hence, we have shown that \lambda(E) \leq \lambda(B). Combining this with \lambda(E) \geq \lambda(B) from your approach, we can conclude that \lambda(E) = \lambda(B). Therefore, B is a G-delta set with E \subseteq B and \lambda(E) = \lambda(B).

I hope this helps. Keep up the good work!