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Fall 2014 - Problem 5

Lebesgue-Radon-Nikodym derivative, measure theory

Let φ ⁣:[0,1]→[0,1]\func{\phi}{\br{0,1}}{\br{0,1}} be continuous and let dμ\diff\mu be a Borel probability measure on [0,1]\br{0,1}. Suppose μ(φ−1(E))=0\mu\p{\phi^{-1}\p{E}} = 0 for every Borel set E⊆[0,1]E \subseteq \br{0,1} with μ(E)=0\mu\p{E} = 0. Show that there is a Borel measurable function w ⁣:[0,1]→[0,∞)\func{w}{\br{0,1}}{\pco{0,\infty}} so that

∫(f∘φ)(x) dμ(x)=∫f(y)w(y) dμ(y)for all continuous f ⁣:[0,1]→R.\int \p{f \circ \phi}\p{x} \,\diff\mu\p{x} = \int f\p{y} w\p{y} \,\diff\mu\p{y} \quad\text{for all continuous } \func{f}{\br{0,1}}{\R}.
Solution.

Observe that μ∘φ−1\mu \circ \phi^{-1} is a measure on [0,1]\br{0, 1}: μ(φ−1(∅))=μ(∅)=0\mu\p{\phi^{-1}\p{\emptyset}} = \mu\p{\emptyset} = 0 and if {En}n\set{E_n}_n is a sequence of disjoint measurable sets, φ−1(En)∩φ−1(Em)=∅\phi^{-1}\p{E_n} \cap \phi^{-1}\p{E_m} = \emptyset if n≠mn \neq m, and so σ\sigma-additivity follows from σ\sigma-additivity of μ\mu.

Notice that the assumption on μ∘φ−1\mu \circ \phi^{-1} says precisely that μ∘φ−1\mu \circ \phi^{-1} is absolutely continuous with respect to μ\mu. Hence, because μ\mu is a finite measure, we may apply the Radon-Nikodym theorem to obtain w∈L1(μ)w \in L^1\p{\mu} such that

μ(φ−1(E))=∫Ew(y) dμ(y).\mu\p{\phi^{-1}\p{E}} = \int_E w\p{y} \,\diff\mu\p{y}.

Notice also that

μ(φ−1(E))=∫χφ−1(E) dμ=∫χE∘φ dμ.\mu\p{\phi^{-1}\p{E}} = \int \chi_{\phi^{-1}\p{E}} \,\diff\mu = \int \chi_E \circ \phi \,\diff\mu.

By linearity, we then see that

∫(f∘φ)(x) dμ(x)=∫f(y)w(y) dμ(y)\int \p{f \circ \phi}\p{x} \,\diff\mu\p{x} = \int f\p{y} w\p{y} \,\diff\mu\p{y}

holds if ff is simple. Since simple functions are dense in L1L^1, it follows that this formula holds for f∈L1f \in L^1 and in particular, for ff continuous.