Final answer:
The composition f \circ \phi is measurable because f is measurable and \phi^{-1} is Lipschitz, implying that \phi^{-1} is continuous. Thus, the pre-image of any measurable set under f \circ \phi is also measurable.
Step-by-step explanation:
To prove that the composition f \circ \phi is measurable, let's begin by establishing what we have:
- f:Rd\u2192R is a measurable function.
- \phi:Rd\u2192Rd is a bijection, meaning it is a one-to-one and onto function.
- The inverse function \phi-1 is Lipschitz, meaning there exists a constant L \u003e 0 such that for all x, y in Rd, the inequality |\phi-1(x) - \phi-1(y)| \u2264 L|x - y| holds.
To show that f \circ \phi is measurable, we need to demonstrate that the pre-image of any open set under f \circ \phi is a measurable subset of Rd. An open set O in R under f has a measurable pre-image since f is measurable. Now consider an open set O under f \circ \phi.
The pre-image of O under f \circ \phi is \phi-1\(f-1\(O)). Because f-1\(O) is measurable and \phi-1\ is Lipschitz, hence continuous, the pre-image of a measurable set under a continuous function is also measurable. Therefore, the composition f \circ \phi satisfies the condition for measurability.