Proof of Radon-Nikodym theorem

Transcription

1 Measure theory class notes - 13 October 2010, class 20 1 Proof of Radon-Niodym theorem Theorem (Radon-Niodym). uppose Ω is a nonempty set and a σ-field on it. uppose µ and ν are σ-finite measures on (Ω, ) such that for all, µ() 0 implies ν() 0. Then There exists z : Ω [0, ) measurable such that for all, ν() z dµ. uch a z is unique upto a.e. equality (w.r.t. µ). z is integrable w.r.t. µ if and only if ν is a finite measure. Proof. ssuming we have obtained the z, it is easy to see that z is integrable if and only if ν is a finite measure, since ν(ω) z dµ. Uniqueness We show uniqueness first. uppose there are two z 1, z 2 : Ω [0, ) such that for all, z 1 dµ ν() z 2 dµ. We would lie to say z 1 dµ z 2 dµ 0, but both the integrals might be. Both µ and ν are σ-finite. Let be a countable partition of Ω such that for all, µ() < and ν() < (to obtain this we can tae such a partition for µ, and such a partition for ν, and tae pairwise intersections). Fix a. Let B {ω : z 1 (ω) z 2 (ω) > 0} Note that 1 B (z 1 z 2 ) 1 (z 1 z 2 ) + (evaluate both sides at ω: for ω /, both are 0; for ω \B, LH is 0 and since (z 1 z 2 )(ω) 0, RH is 0; for ω B both side are z 1 (ω) z 2 (ω)). ν(b) 1B z 1 dµ 1 B z 2 dµ. ince ν(b) ν() <, it is meaningful to consider 1 B z 1 dµ 1 B z 2 dµ. 0 1 B z 1 dµ 1 B z 2 dµ 1 B (z 1 z 2 ) dµ 1 (z 1 z 2 ) + dµ If the integral of a nonnegative function is 0, it must be 0 almost everywhere. pplying to the above: µ({ω : ω and z 1 (ω) z 2 (ω) > 0}) 0 We can interchange the roles of z 1 and z 2 in the above argument. ombining the two, we get µ({ω : ω and z 1 (ω) z 2 (ω) 0}) 0 This holds for all. ince is countable, and a countable union of null sets is a null set, µ({ω : z 1 (ω) z 2 (ω)}) 0 This proves that the z promised in the theorem (if it exists) is unique upto a.e. (w.r.t. µ) equality.

2 Measure theory class notes - 13 October 2010, class 20 2 Existence We now show the existence of such a z. We reduce the general case to the case when both µ and ν are finite, and then show existence for the finite case. Reduction to the finite measure case uppose we now the Radon-Niodym theorem holds for the case when the measures involved are finite. In general, assume µ and ν are σ-finite, and let be a countable partition of Ω such that for all, µ() and ν() are finite. Define finite measures µ and ν for all, µ () µ( ), ν () ν( ) If µ () 0, then ν () 0. Using the finite-measure case of the theorem, let z : Ω [0, ) be such that for all and, ν () z dµ z and z 1 differ at most on Ω \, which has 0 µ measure. o we can replace z by z 1, and so assume that for ω /, z (ω) 0. Define z : Ω [0, ), by requiring that z (z ) (that is, for ω, z(ω) z (ω)). z can also be described as z - this shows that z is measurable. If g : Ω [0, ) is 0 outside, then g dµ g dµ. This can be proven as usual by proving it for simple functions and then for all nonnegative measurable functions. This fact is applied to z in what follows. For any, ν() ν( ) ν () z dµ z dµ z dµ z dµ It remains to prove the theorem for the finite measure case: (by taing partial sums; MT) Finite measures - getting hold of the z: Now assume µ and ν are finite measures. Let { f : (f : Ω [0, ] R), ( f dµ ν())}

3 Measure theory class notes - 13 October 2010, class 20 3 We want to identify f such that equality holds, that is, for all, f dµ ν(). is nonempty, because the zero function belongs to. Intuitively, we want to choose the largest function in as a candidate for our z. Does such a largest function exist? We observe some properties of first: If f, g, then f g (which is the pointwise maximum of f and g) belongs to. Tae any. We apply the condition defining to the sets 1 : {ω : f(ω) g(ω)} and 2 : {ω : f(ω) < g(ω)}: 1 f ν( 1 ), 2 g ν( 2 ) Note that f1 1 (f g)1 1 and g1 2 (f g)1 2. ubstituting in the above equations and adding them, we get (f g) ν() uppose {f n } n1 is an increasing sequence in. Let f be their supremum (equivalently limit). Then f : tae any. {f n 1 } n1 increases to f1, and so by the monotone convergence theorem the corresponding limit with integrals holds. ince each f n 1 ν(), we have f1 ν(). We identify the largest function in by considering the function with the largest integral. For all f, f dµ ν(ω) <. Let { } α sup f dµ : f 0 α ν(ω). In particular, α. Let {f n } n1 be a sequence from such that { f n dµ } n1 converges to α (this can be done because of the supremum definition of α). onsider the partial maxima of {f n } n1, that is, define {g n } n1 by g n max{f 1, f 2,..., f n } We now that g n, so g n dµ α. lso, f n g n, so f n dµ g n dµ α. Thus lim n gn dµ α. {g n } n1 is an increasing sequence; let z lim n g n. By the monotone convergence theorem, z dµ α. ince z is integrable, z is finite almost everywhere. We modify z on a set of measure 0 suitably so that z never taes the value. This z is our largest function in, as we show now: Let g. We will show that g z a.e. (w.r.t µ). Let {ω : g(ω) > z(ω)}. Let z g h. h. {ω : h(ω) > z(ω)} {ω : h(ω) z(ω) > 0}. h z; h dµ z dµ α, so h α. h z is a nonnegative function whose µ integral is 0, so h z is 0 almost everywhere. µ() 0. g z almost everywhere (w.r.t µ).

4 Measure theory class notes - 13 October 2010, class 20 4 Finite measures - showing that the z wors Define λ : [0, ), λ() ν() Note that this is well-defined, that is, the expression on the RH indeed belongs to [0, ). λ is a measure: clearly λ( ) 0. Let be a countable collection of disjoint sets. ν() + ( ) z dµ ν + z dµ < ince the relevant series is absolutely convergent, we can rearrange terms, and ( ) ( ) λ ν z dµ ν() z dµ ( ) ν() z dµ λ() z dµ λ is a measure. We want to show that λ is the zero measure, and then we will be done. If for all and all N, λ() 1 µ() then since µ() is finite, λ() 0, and we are done. o there exists and N such that λ() 1 µ() > 0. Fix such an and. We will derive a contradiction. all a set Z to be good if the following two conditions hold: λ(z) 1 µ(z) > 0. For all B Z with B, λ(b) 1 µ(b) 0. The first condition implies that a good set Z must have µ(z) > 0: otherwise if µ(z) 0, then by the hypothesis, ν(z) 0, and so λ(z) 0. satisfies the first condition, but need not satisfy the second condition. We will identify a good set - itself may not be a good set, but we can scrape off some parts of it to get a good set. Once we identify a good set Z, we will use the function z Z to get a contradiction. Let 0. Let β 0 inf{λ(b) 1 µ(b) : B, B }. If β 0 0, then itself is a good set. o assume β 0 < 0. β 0 cannot be, since it is at least (ν(ω) + z dµ + 1 µ(ω)). Pic B 0 0 such that λ(b 0 ) 1 µ(b 0) < 1 2 β 0 Put 1 \ B 0. λ( 1 ) 1 µ( 1) ( λ() λ(b 0 ) ) ( 1 µ() 1 µ(b 0) ) λ() 1 µ() > 0 (because λ(b 0 ) 1 µ(b 0) < 1 2 β 0 < 0.) n important observation is this: if B 1, then λ(b) 1 µ(b) 1 2 β 0. Otherwise, we could tae a B which violates this, and B B 0 would violate the infimum definition of β 0. If 1 is a good set, we have found a good set. Otherwise, repeat with 1 what we did with 0. The above observation says that the β 1 so obtained will satisfy β β 0. Inductively, we construct a sequences of sets { n } n1 and {B n } n1, and a sequence of real numbers {β n } n1, such that the following holds: some n is good, or for all n N,

5 Measure theory class notes - 13 October 2010, class 20 5 B n n. n+1 n \ B n. β n inf { λ(b) 1 µ(b) : B n, B }, β n < 0. λ( n+1 ) 1 µ( n+1) λ( n ) 1 µ( n). β n β n. This immediately implies that for all n, β n 1 β 2 n 0 and λ( n ) 1µ( n) λ() 1 µ() > 0. If no n was good, define n N n. ince λ and µ are finite measures, λ( ) lim n λ( n ) and µ( ) lim n µ( n ) o λ( ) 1µ( ) λ() 1µ() > 0. If B and B, then since B n, λ(b) 1µ(B) 1 β 2 n 0. ince this holds for all n, λ(b) 1µ(B) 0. is a good set! If some n was good, let be that good set. We will show that z For any, if Ω \, then If, then (z ) dµ z dµ ν() ( z ) dµ z dµ + 1µ() z dµ + λ() (since is a good set) ν() (by definition of λ) For general, we just combine the part of in and the part in Ω \ : (z + 1 ) 1 dµ (z + 1 ) 1 dµ+ (z + 1 ) 1 dµ ν( )+ν( \) ν() \ We have shown that z ince is a good set, µ( ) > 0. It is not the case that z z a.e. (w.r.t. µ), which contradicts what we have shown earlier. λ must be the zero measure, and so for all D, ν(d) z dµ D