Today's post is about the Monotone Convergence Theorem and the theorems leading up to it.
Theorem (itg2mono): If (fn)n∈N is a sequence of measurable functions R→[0,∞) such that 0≤f1≤f2≤⋯≤fn≤… and the pointwise limit g of the sequence (fn) exists, then g is measurable and ∫g=limn→∞∫fn (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}={x∈R|g(x)>t} is measurable, which follows from {g>t}=⋃n∈N{fn>t}, because t<g(x)=supn∈Nfn(x)⟺∃n∈N:t<fn(x) and the countable union of the measurable sets {fn>t} is measurable.
Before tackling the main claim ∫g=limn→∞∫fn, we will need a few more lemmas, so let's get those down first.
Theorem (volsup): If (An)n∈N is a sequence of measurable sets such that A1⊆A2⊆⋯⊆An⊆⋯⊆R and ⋃n∈NAn=A, then vol(A)=limn→∞vol(An), where the limit may be ∞.
Proof: First, we address the case when there is an Ak such that vol(Ak)=∞. In this case, Ak⊆A implies vol(A)=∞, and limn→∞vol(An)=supn→∞vol(An)≥vol(Ak)=∞ so limn→∞vol(An)=∞ as well.
Otherwise, all the vol(An) are real, so we are free to do arithmetic on the volumes. Consider the sequence Bn=An∖⋃n−1k=1Ak=An∖An−1 (where the second representation is only valid for n≥2). This is a set of pairwise-disjoint measurable sets, so from voliun we have vol(⋃n∈NBn)=∑∞n=1vol(Bn). Now each Bn⊆An, so ⋃n∈NBn⊆A, and each x∈A is in some Ak, and for the minimal k, it is also in Bk. Thus A=⋃n∈NBn. 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)=limn→∞vol(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+1∑k=1vol(Bn)=vol(An)+vol(Bn+1)=vol(An)+vol(An+1∖An), and since these are disjoint sets and An⊆An+1 we have vol(An)+vol(An+1∖An)=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)n∈N is a sequence of measurable sets such that A1⊆A2⊆⋯⊆An⊆⋯⊆R and ⋃n∈NAn=R, and ϕ is a simple function, then limn→∞∫Anϕ=∫ϕ.
Proof: We expand out the definition of the integral on simple functions: ∫ϕ=∑x∈R∗xvol(ϕ−1{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: limn→∞∫Anϕ=limn→∞∑x∈R∗xvol(ϕ−1{x}∩An)=∑x∈R∗xlimn→∞vol(ϕ−1{x}∩An), so we are reduced to proving limn→∞vol(ϕ−1{x}∩An)=vol(ϕ−1{x}), which is an application of volsup above, taking the sets (ϕ−1{x}∩An)n∈N 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 limn→∞∫fn≤∫g, which is true because each fn≤g so ∫fn≤∫g by itg2le. Now the definition of ∫g is as a supremum over all simple functions ϕ≤g. Thus to show ∫g≤limn→∞∫fn we can show that S:=limn→∞∫fn is an upper bound on ∫ϕ for any ϕ≤g. Now we are going to prove that t∫ϕ≤S for any 0<t<1, but to get from there to ∫ϕ≤S takes a little trickery.
We assume that ∫ϕ>S to start with. Since S∈[0,∞] and ∫ϕ∈R 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 S∈R and ∫ϕ>0. Now we can use alrple to use that for any x>0, ∫ϕ≤S+x and hence prove that ∫ϕ≥S (in contradiction to our assumption, thus showing ∫ϕ≥S unconditionally). Define t=max(12,SS+x), which is valid since S≥0 and x>0. Then t≥12>0 and t≤SS+x<1, so by our lemma below, t∫ϕ≤S. Since ∫ϕ>0 and t>0 we can infer S>0, so we can cancel the S and multiply by S+x in SS+x∫ϕ≤t∫ϕ≤S to get ∫ϕ≤S+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 ϕ≤g, we have ∫ϕ≤limn→∞∫fn.
Proof: Define the sequence An={x∈R|tϕ(x)≤fn(x)}. Then ∫Antϕ≤∫fn because the restricted function tϕ↾An is always dominated by fn: if x∉An then (tϕ↾An)(x)=0≤f(n) and if x∈An then (tϕ↾An)(x)=tϕ(x)≤f(n). We can take this inequality in the limit, and thus ∫tϕ=limn→∞∫Antϕ≤limn→∞∫fn, where the first equality is none other than our earlier lemma itg1climres, which can be applied because An={0≤fn−tϕ} is a measurable set (since fn−tϕ is a sum of a simple function and a measurable function, which is measurable), the An sets form an inclusion chain since fn+1≤fn, 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)≤0≤fn(x) for every n. This completes the proof.
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