We've already noted that $\emptyset$, $\mathbb{R}$, every countable set, and every interval is a Lebesgue measurable set.We now give a more general definition of the collection of Lebesgue measurable sets. Theorem. The complement of an Fσ set is a Gδ set (and conversely). Show that there is a G set Gthat contains Efor which m(E) = m(G); while m(G˘E) >0 Proof. Notice. A further investigation of this topic is left as an exercise. We say that is measurable if the preimage of every measurable set in is measurable. Therefore, for all $\alpha \in \mathbb{R}$ we have that the following sets are Lebesgue measurable: (4) Problem 33. for every open set G ⊆ R. Show that the concepts of measurability in Definition 4.1.1 for both real and complex valued functions are consistent with the concept of measurability for a topological space valued function. Remark 1 Lebesgue measure µ(E) satisfies the properties (1)–(4) on the collection M of measurable subsets of R. However, not all subsets of R are measurable. Every open interval is Fσ: (a,b) = [∞ n=1 a+ 1 n,b − 1 n (a and b could be ±∞), and hence every open set is F σ (this is Problem 1.37). MEASURABLE FUNCTIONS In that case, it follows from Proposition 3.2 that f: X!Y is measurable if and only if f 1(G) 2Ais a measurable subset of Xfor every set Gthat is open in Y.In particular, every continuous function between topological spaces that are equipped We have already looked at whether certain sets are Lebesgue measurable. But for every open set that covers the rationals in, say, [tex] [0,1] [/tex] must cover the entire interval so that the set of rationals can't satisfy the conditions for a measurable set. M 6= 2R. Okay okay, the last one isn't really a fact, but it may not surprise you that the Cantor set is central to today's discussion.

Every set in $\mathscr{L}$ with positive measure contains a non (Lebesgue) measurable subset. 34 3. When proving measurability, there are another set of tools which are useful.

97.3% of all counterexamples in real analysis involve the Cantor set. What is interesting with this definition is it's strong relation to the defintion of continuity between topological spaces, which is, the preimage of every open set is open.

Theorem 3 Properties (1)–(4) imply there exist nonmeasurable sets. Then open interval $(-\infty, \alpha)$ is an open set for all $\alpha \in \mathbb{R}$. The Collection of Lebesgue Measurable Sets.
Definition 3 If E is a Lebesgue measurable set, then the Lebesgue measure of E is ... Every open set and every closed set is measurable. (Problem 1.56) Let f be a real valued function defined on all of R. The set of points at which f is continuous is a Gδ set. Let Ebe a non-measurable set of nite outer measure. But then, this set of rationals in the unit interval is the countable union of point sets so it MUST BE measurable. Note. Young’s Theorem.