Differences

This shows you the differences between two versions of the page.

--- medida:gpt [2023/03/27 16:37] – created - external edit 127.0.0.1
+++ medida:gpt [2023/04/12 19:15] (current) – external edit 127.0.0.1
@@ Line 2: / Line 2: @@
 Read this page with moderation. To detect the gaps/errors in the answers to math questions by ChatGPT typically is very easy. However, I am trying to find some questions/answers where finding the flaw/error can be an exercise for the students. Well, not an easy task at the moment with the free version.
 </WRAP>
+<WRAP center round info 60%>
+prove that the limit of a sequence of subsets En is equal to the set E if and only if the limit of the caracteristic functions of En converges pointwise to the characteristic function of E.
+</WRAP>
+To prove the statement, we need to show two things:
+If the limit of the sequence of subsets En is equal to the set E, then the limit of the characteristic functions of En converges pointwise to the characteristic function of E.
+If the limit of the characteristic functions of En converges pointwise to the characteristic function of E, then the limit of the sequence of subsets En is equal to the set E.
+Here is the proof:
+Suppose that lim En = E. Let χ_A(x) denote the characteristic function of a set A, that is, χ_A(x) = 1 if x ∈ A and χ_A(x) = 0 if x ∉ A. We need to show that lim χ_En(x) = χ_E(x) for all x ∈ X.
+Fix x ∈ X. If x ∈ E, then x belongs to En for infinitely many values of n. Therefore, χ_En(x) = 1 for infinitely many values of n, and hence lim χ_En(x) = 1 = χ_E(x). On the other hand, if x ∉ E, then x does not belong to En for all sufficiently large n. Therefore, χ_En(x) = 0 for all sufficiently large n, and hence lim χ_En(x) = 0 = χ_E(x). This shows that lim χ_En(x) = χ_E(x) for all x ∈ X, as required.
+Now suppose that lim χ_En(x) = χ_E(x) for all x ∈ X. We need to show that lim En = E.
+First, note that if x ∈ E, then χ_E(x) = 1, and hence χ_En(x) = 1 for infinitely many values of n. Therefore, x ∈ En for infinitely many values of n, which implies that x ∈ lim En. On the other hand, if x ∉ E, then χ_E(x) = 0, and hence χ_En(x) = 0 for all sufficiently large n. Therefore, x ∉ En for all sufficiently large n, which implies that x ∉ lim En.
+Thus, we have shown that lim En is a subset of E, and that every element of E is also an element of lim En. This means that lim En = E, as required.
+Therefore, we have shown both directions of the "if and only if" statement, and hence the statement is proved.
@@ Line 67: / Line 92: @@
 Hence, we have shown that E is Lebesgue measurable if and only if E is a G delta set with a null set removed.
+~~DISCUSSIONS~~