p. 53, #37 We will do this in three problems: Note here I want you to assume f: R→ R is Lipschitz 3) Prove that if S CR has measure zero then so does f(S). Hint: For 0 pick a cover of A by finite open intervals whose sum of lengths is smaller than e. Now naturally form a cover of f(A) by closed intervals whose sum of lengths is also small. Now use countable subadditivity! 4) Prove that if A C R is F。 then so is f(A). Hint: This just requires f to be con- tinuous! Note that A' closed doesn't necessarily mean that f(A') is closed, BUT A' compact implies f(A') is compact. 5) Using Theorem 11 on p. 40 (i.e. E measurable iff there exists Fo set F C E where m*(E\F) = 0, prove that E measurable implies f(E) measurable. NOTE: this is FALSE for f RR continuous! See p. 52.

Fig: 1