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Sunday, August 31, 2014

Integral of a positive function is positive

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 xA, 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 0xvol(A), then there is a measurable BA 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 nN. This is an increasing sequence of measurable sets, so by volsup we have supnNvol(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 yy, 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])=yy, so f is a continuous function RR. Since f(n)=0x<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 xR there is a BA with vol(B)=x/c, and then Bcdx=xAcdx. 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 xA so that there is some k with f(x)>1k, AnNAn. Now assume that Af(x)dx=0. Then 0An1ndxAnf(x)dxAf(x)dx=0, so from itg2const and itg2const2, vol(An)=0 (because vol(An)= implies An1ndx=0 and vol(An)R implies nAn1ndx=vol(An)=0). Since this is true for every n, we have vol(A)vol(nNAn)=limnvol(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.

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