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Saturday, August 16, 2014

The Monotone Convergence Theorem for integrals

Today's post is about the Monotone Convergence Theorem and the theorems leading up to it.

Theorem (itg2mono): If (fn)nN is a sequence of measurable functions R[0,) such that 0f1f2fn and the pointwise limit g of the sequence (fn) exists, then g is measurable and g=limnfn (note that for increasing sequences, lim and sup are equivalent, so the statement may differ from the Metamath version for this reason).

Proof: First, we prove g is measurable. (mbfmono) It is sufficient to prove that the inverse image {g>t}={xR|g(x)>t} is measurable, which follows from {g>t}=nN{fn>t}, because t<g(x)=supnNfn(x)nN:t<fn(x) and the countable union of the measurable sets {fn>t} is measurable.

Before tackling the main claim g=limnfn, we will need a few more lemmas, so let's get those down first.

Theorem (volsup): If (An)nN is a sequence of measurable sets such that A1A2AnR and nNAn=A, then vol(A)=limnvol(An), where the limit may be .

Proof: First, we address the case when there is an Ak such that vol(Ak)=. In this case, AkA implies vol(A)=, and limnvol(An)=supnvol(An)vol(Ak)= so limnvol(An)= as well.

Otherwise, all the vol(An) are real, so we are free to do arithmetic on the volumes. Consider the  sequence Bn=Ann1k=1Ak=AnAn1 (where the second representation is only valid for n2). This is a set of pairwise-disjoint measurable sets, so from voliun we have vol(nNBn)=n=1vol(Bn). Now each BnAn, so nNBnA, and each xA is in some Ak, and for the minimal k, it is also in Bk. Thus A=nNBn. As for the sum, we prove by induction that nk=1vol(Bn)=vol(An) and therefore the sequence of partial sums of vol(Bn) is the sequence vol(An) and n=1vol(Bn)=limnvol(An).

For n=1, we have B1=A1 so 1k=1vol(Bn)=vol(A1), and assuming that this is true for n, we have n+1k=1vol(Bn)=vol(An)+vol(Bn+1)=vol(An)+vol(An+1An), and since these are disjoint sets and AnAn+1 we have vol(An)+vol(An+1An)=vol(An+1).

Next we use this theorem to prove that the integral on simple functions respects limit operations on the base set.

Theorem (itg1climres): If (An)nN is a sequence of measurable sets such that A1A2AnR and nNAn=R, and ϕ is a simple function, then limnAnϕ=ϕ.

Proof: We expand out the definition of the integral on simple functions: ϕ=xRxvol(ϕ1{x}), where the sum is nominally over all nonzero reals but is only nonzero for finitely many x. We can distribute the limit over the sum: limnAnϕ=limnxRxvol(ϕ1{x}An)=xRxlimnvol(ϕ1{x}An), so we are reduced to proving limnvol(ϕ1{x}An)=vol(ϕ1{x}), which is an application of volsup above, taking the sets (ϕ1{x}An)nN as our increasing sequence.

Now we are finally ready to return to our proof of the main theorem itg2mono. The easy direction of our equality is limnfng, which is true because each fng so fng by itg2le. Now the definition of g is as a supremum over all simple functions ϕg. Thus to show glimnfn we can show that S:=limnfn is an upper bound on ϕ for any ϕg. Now we are going to prove that tϕS for any 0<t<1, but to get from there to ϕS takes a little trickery.

We assume that ϕ>S to start with. Since S[0,] and ϕR to start with (remember that S is a supremum, possibly infinite, of integrals of nonnegative functions, while ϕ is a simple function whose integral is real but possibly negative), the inequality ϕ>S immediately implies SR and ϕ>0. Now we can use alrple to use that for any x>0, ϕS+x and hence prove that ϕS (in contradiction to our assumption, thus showing ϕS unconditionally). Define t=max(12,SS+x), which is valid since S0 and x>0. Then t12>0 and tSS+x<1, so by our lemma below, tϕS. Since ϕ>0 and t>0 we can infer S>0, so we can cancel the S and multiply by S+x in SS+xϕtϕS to get ϕS+x as we wanted to show.

Now all that remains to prove is our lemma:

Lemma (itg2monolem): For any 0<t<1, and any simple function ϕg, we have ϕlimnfn.

Proof: Define the sequence An={xR|tϕ(x)fn(x)}. Then Antϕfn because the restricted function tϕAn is always dominated by fn: if xAn then (tϕAn)(x)=0f(n) and if xAn then (tϕAn)(x)=tϕ(x)f(n). We can take this inequality in the limit, and thus tϕ=limnAntϕlimnfn, where the first equality is none other than our earlier lemma itg1climres, which can be applied because An={0fntϕ} is a measurable set (since fntϕ is a sum of a simple function and a measurable function, which is measurable), the An sets form an inclusion chain since fn+1fn, and the union of all of them is R because either ϕ(x)>0 in which case tϕ(x)<ϕ(x)g(x) and there is some fn(x)>tϕ(x), or ϕ(x)0 in which case tϕ(x)0fn(x) for every n. This completes the proof.

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