Here is a nice theorem that is not as simple as the statement would have you believe:
Theorem (itggt0): If f is integrable and A is a measurable set such that 0<vol(A)≤∞ and f(x)>0 for all x∈A, then ∫Af(x)dx>0.
The analogous theorem with ≤ is trivial (itgge0), but this theorem requires some of the countable additivity properties of the volume function. We postpone the proof until we have proven the needed lemmas. First, we establish what I will call, for lack of a better name, the Intermediate Value Theorem for the Lebesgue volume function.
Theorem (volivth): If A is a measurable set and 0≤x≤vol(A), then there is a measurable B⊆A such that vol(B)=x. (Note that vol(A) is allowed to be infinite.)
Proof: If x=vol(A), then B=A satisfies the requirements, so assume instead that x<vol(A). Consider the sequence A∩[−n,n] for n∈N. This is an increasing sequence of measurable sets, so by volsup we have supn∈Nvol(A∩[−n,n])=vol(A). Thus since x<vol(A) there is an n such that x<vol(A∩[−n,n]).
Now define the function f(y)=vol(A∩[−n,y]). If y≤y′, then [−n,y]⊆[−n,y′]⟹f(y)≤f(y′), so f is a weakly increasing function, and since (A∩[−n,y′])∖(A∩[−n,y])⊆[y,y′], we also have f(y′)−f(y)≤vol([y,y′])=y′−y, so f is a continuous function R→R. Since f(−n)=0≤x<vol(A∩[−n,n])=f(n), the regular intermediate value theorem implies that there is some z such that vol(A∩[−n,z])=x, and for this z, B=A∩[−n,z] satisfies the requirements of the conclusion.
We need this theorem to extend the formula ∫Acdx=cvol(A) to the case vol(A)=∞. (We can't use itg2const directly because the formula is only valid when vol(A) is a real number, since multiplication only works on complex numbers.) If vol(A)=∞ and c>0, then for any x∈R there is a B⊆A with vol(B)=x/c, and then ∫Bcdx=x≤∫Acdx. Thus ∫Acdx is larger than any real number, so ∫Acdx=∞ as well. (This is theorem itg2const2.)
Now we can prove the main theorem.
Proof (of itggt0): Define the sequence An={x:f(x)>1n}. This is clearly an increasing sequence of measurable sets (since f is measurable), and since f(x)>0 for all x∈A so that there is some k with f(x)>1k, A⊆⋃n∈NAn. Now assume that ∫Af(x)dx=0. Then 0≤∫An1ndx≤∫Anf(x)dx≤∫Af(x)dx=0, so from itg2const and itg2const2, vol(An)=0 (because vol(An)=∞ implies ∫An1ndx=∞≠0 and vol(An)∈R implies n∫An1ndx=vol(An)=0). Since this is true for every n, we have vol(A)≤vol(⋃n∈NAn)=limn→∞vol(An)=0, which contradicts the hypothesis vol(A)>0. Therefore ∫Af(x)≠0, and since ∫Af(x)≥0 we in fact have ∫Af(x)>0.
No comments:
Post a Comment