The Radon-Nikodym Theorem Yongheng Zhang Theorem 1 (Johann Radon-Otton Nikodym). Let (X; B; µ) be a σ-finite measure space and let ν be a measure defined on B such that ν µ. Then there is a unique nonnegative measurable function f up to sets of µ-measure zero such that Z ν(E) = fdµ, E for every E 2 B. f is called the Radon-Nikodym derivative of ν with respect to µ and it is often dν denoted by [ dµ ]. The assumption that the measure µ is σ-finite is crucial in the theorem (but we don't have this requirement directly for ν). Consider the following two examples in which the µ's are not σ-finite. Example 1. Let (R; M; ν) be the Lebesgue measure space. Let µ be the counting measure on M. So µ is not σ-finite. For any E 2 M, if µ(E) = 0, then E = ; and hence ν(E) = 0. This shows ν µ. Do we have the associated Radon-Nikodym derivative? Never. Suppose we have it and call it R R f, then for each x 2 R, 0 = ν(fxg) = fxg fdµ = fxg fχfxgdµ = f(x)µ(fxg) = f(x). Hence, f = 0. R This means for every E 2 M, ν(E) = E 0dµ = 0, which contradicts that ν is the Lebesgue measure. Example 2. ν is still the same measure above but we let µ to be defined by µ(;) = 0 and µ(A) = 1 if A 6= ;. Clearly, µ is not σ-finite and ν µ. The Radon-Nikodym derivative does not exist, neither. R Suppose f is one. Then for any x 2 R, 0 = ν(fxg) = fxg fdµ = f(x)µ(fxg) = f(x)1. Thus, f = 0. Hence for any E 2 M, ν(E) = infE 0dµ = 0. [Also notice that if µ(X) = 0, then the measure of every set in M with respect to µ and ν is zero, which is not interesting at all. So we ignore this case.] Now we prove the Radon-Nikodym Theorem. Proof(Wikipedia): We assume that (X; B; µ) is a finite measure space and show the existence and uniqueness of the Radon-Nikodym derivative. Then the case when µ is σ-finite follows by the \past- ing" of the derivatives on each set of finite measure. Let F = ff is measurablejν(E) ≥ R fdµ, for all E 2 Mg. F is nonempty because the zero func- R E R tion is in it. Let s = supf2F X fdµ. Then there is a sequence hhni in F such that limn!1 X fndµ = s. R R R Let f1; f2 2 F, then for any E 2 M, E f1_f2dµ = fx2Ejf (x)≥f (x)g f1_f2dµ+ fx2Ejf (x)<f (x)g f1_ R R 1 2 1 2 f2dµ = f1dµ + f2dµ ≤ ν(fx 2 Ejf1(x) ≥ f2(x)g) + ν(fx 2 fx2Ejf1(x)≥f2(x)g fx2Ejf1(x)<f2(x)g Ejf1(x) < f2(x)g) = ν(E). Therefore, f1 _ f2 2 F. Wn R Let fn = k=1 hk. Then hfni is a nonnegative increasing sequence in F and limn!1 X fndµ = s. Define g by g(x) = limn!1 fn(x) for x 2 X. Then by the motonone convergence theorem, for any R R R R E 2 M, E gdµ = limn!1 E fndµ ≤ ν(E). This shows g 2 F and X gdµ = limn!1 X fndµ = s. 1 2 R Therefore, the function ν0 defined on M by ν0(E) = ν(E) − E gdµ is a measure. We want to show that ν0 = 0 and then g is the desired function. Suppose ν0 is not zero. Since ν0(X) > 0 and µ(X) < 1, there is > 0 such that ν0(X) − µ(X) > 0. Let fA; Bg be a Hahn decompo- sition for the signed measure ν0 − µ. Then for every E 2 M, ν0(A \ E) − µ(A \ E) ≥ 0. So, R R R R ν(E) = ν0(E)+ gdµ ≥ ν0(E \A)+ gdµ ≥ µ(A\E)+ gdµ = (g +χA). Therefore, g +χA E ER R E E R is also in F. However, if µ(A) > 0, then (g+χA)dµ = gdµ+µ(A) > gdµ = s, which is a con- X X RX tradiction. In fact, if µ(A) = 0, since ν µ, ν(A) = 0. So ν0(A) = ν(A) − A gdµ ≤ ν(A) = 0. Hence ν0(A) = 0. Consequently, ν0(X)−µ(X) = ν0(B)−µ(B) ≤ 0, contradicting that ν0(X)−µ(X) > 0. R Therefore, ν0 = 0, which means ν(E) = E gµ for every E 2 M. To show uniqueness, let ν(E) = R fdµ = R gdµ. Then R (f − g)dµ = 0. Since E is arbitrary, R E E E ff−g≥0g f − gdµ = 0. This shows f = g a.e. on fx 2 Xjf(x) ≥ g(x)g. Similarly, f = g a.e. on fx 2 Xjf(x) < g(x)g. Hence f = g a.e. on X. Remark: If we add the condition that ν is finite, then the function g in our proof is integrable..