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In fact, Φ is continuous, so Φ−1 ((−∞, a)) is an open set O, and hence (Φ ◦ f )−1 ((−∞, a)) = f −1 (O) is measurable. It should be noted, however, that in general it is not true that f ◦ Φ is measurable whenever f is measurable and Φ is continuous. See Exercise 35. Property 3 Suppose {fn }∞ n=1 is a sequence of measurable functions. Then sup fn (x), n inf fn (x), lim sup, fn (x) n and n→∞ lim inf fn (x) n→∞ are measurable. Proving that supn fn is measurable requires noting that {supn fn > a} = n {fn > a}.

If we let FK = j=K+1 Ej and F = K=1 FK , then m(F ) = 0 since m(FK ) ≤ 2−K , and ψk (x) → f (x) for all x in the complement of F , which is the desired result. 33 4. 3 Littlewood’s three principles Although the notions of measurable sets and measurable functions represent new tools, we should not overlook their relation to the older concepts they replaced. Littlewood aptly summarized these connections in the form of three principles that provide a useful intuitive guide in the initial study of the theory.

7. If δ = (δ1 , . . , δd ) is a d-tuple of positive numbers δi > 0, and E is a subset of Rd , we deﬁne δE by δE = {(δ1 x1 , . . , δd xd ) : where (x1 , . . , xd ) ∈ E}. Prove that δE is measurable whenever E is measurable, and m(δE) = δ1 · · · δd m(E). 8. Suppose L is a linear transformation of Rd . Show that if E is a measurable subset of Rd , then so is L(E), by proceeding as follows: (a) Note that if E is compact, so is L(E). Hence if E is an Fσ set, so is L(E). (b) Because L automatically satisﬁes the inequality |L(x) − L(x )| ≤ M |x − x | for some M , we can see that L maps√any cube of side length into a cube of side length cd M , with cd = 2 d.