Write E 1 H 1 Z 1 H 1 is G \u03b4 set and Z 1 0 The same for E 2 H 2 Z 2 Then E 1 E

# Write e 1 h 1 z 1 h 1 is g δ set and z 1 0 the same

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Write E 1 = H 1 Z 1 , H 1 is G δ set and | Z 1 | = 0 . The same for E 2 = H 2 Z 2 . Then E 1 × E 2 = H 1 × H 2 Z 1 × H 2 H 1 × Z 2 where H 1 × H 2 = Ã \ n =1 O n ! × Ã \ n =1 ˜ O n ! = \ n =1 ³ O n × ˜ O n ´ is a G δ set in R 2 ( O n and ˜ O n are open set in R 1 ). We see that H 1 × H 2 R 2 is measurable. Assume fi rst that | H 2 | < . One can fi nd open sets G 1 Z 1 , G 2 H 2 such that | G 1 | < ε, | G 2 | < | H 2 | + ε. Write G 1 = [ j =1 I j (nonoverlapping closed intervals in R 1 ) G 2 = [ k =1 ˜ I k (nonoverlapping closed intervals in R 1 ) 1
to see that Z 1 × H 2 G 1 × G 2 = [ j, k =1 ³ I j × ˜ I k ´ (nonoverlapping closed intervals in R 2 ) with | G 1 × G 2 | = X j,k ¯ ¯ ¯ I j × ˜ I k ¯ ¯ ¯ = X | I j | · X ¯ ¯ ¯ ˜ I k ¯ ¯ ¯ = | G 1 | · | G 2 | < ε ( | H 2 | + ε ) (0.3) which implies | Z 1 × H 2 | = 0 . If | H 2 | = , decompose H 2 = S n = −∞ ( H 2 [ n, n + 1)) , disjoint union, and see that | Z 1 × H 2 | = 0 . The same for | H 1 × Z 2 | = 0 . Hence E 1 × E 2 R 2 is measurable. Next we show that | E 1 × E 2 | = | E 1 | × | E 2 | Case 1: | E 1 | < , | E 2 | < . From (0.3) we see that | G 1 × G 2 | = | G 1 | × | G 2 | for any two open sets in R 1 .

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• Fall '08
• Akhmedov,A
• Algebra, Topology, Sets, Empty set, Metric space, Open set, Topological space, Closed set

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