By Tao T.

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Since E is null for m, it is null for mf and µ also, and so µs is trivial, giving (i). 6 (Lebesgue decomposition theorem). Let m be an unsigned σ-finite measure, and let µ be a signed σ-finite measure. Then there is a unique decomposition µ = µac + µs , where µac m and µs ⊥ m. ) If µ is unsigned, then µac and µs are also. 9. If every point in X is measurable, we call a signed measure µ continuous if µ({x}) = 0 for all x. 6, but suppose also that every point is measurable and m is continuous. Show that there is a unique decomposition µ = µac + µsc + µpp , where µac m, µpp is supported on an at most countable set, and µsc is both singular with respect to m and continuous.

21), we obtain the claim. In the p = 1 case, we instead use f := 1g>N as the test functions, to conclude that g is bounded almost everywhere by N ; we leave the details to the reader. This handles the case when µ is finite. When µ is σ-finite, we can write X as the union of an increasing sequence En of sets of finite measure. On each such set, the above arguments let us write λ = λgn for some gn ∈ Lp (En ). The uniqueness arguments tell us that the gn are all compatible with each other, in particular if n < m, then gn and gm agree on En .

If one already has a metric, then one of course has a topology generated by the open balls of that metric; but there are many important topologies on function spaces in analysis that do not arise from metrics. We also often require the topology to be compatible with the other structures on the function space; for instance, we usually require the vector space operations of addition and scalar multiplication to be continuous. In some cases, the topology on V extends to some natural superspace W of more general functions that contain V ; in such cases, it is often 3This will be the only type of metric on function spaces encountered in this course.