## 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 $0<\DeclareMathOperator{\vol}{vol}\vol(A)\le\infty$ and $f(x)>0$ for all $x\in A$, then $\int_A f(x)\,dx>0$.

The analogous theorem with $\le$ 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 $0\le x\le\vol(A)$, then there is a measurable $B\subseteq A$ 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 $A\cap[-n,n]$ for $n\in\Bbb N$. This is an increasing sequence of measurable sets, so by volsup we have $\sup_{n\in\Bbb N}\vol(A\cap[-n,n])=\vol(A)$. Thus since $x<\vol(A)$ there is an $n$ such that $x<\vol(A\cap[-n,n])$.

Now define the function $f(y)=\vol(A\cap[-n,y])$. If $y\le y'$, then $[-n,y]\subseteq[-n,y']\implies f(y)\le f(y')$, so $f$ is a weakly increasing function, and since $(A\cap[-n,y'])\setminus(A\cap[-n,y])\subseteq[y,y']$, we also have $f(y')-f(y)\le\vol([y,y'])=y'-y$, so $f$ is a continuous function $\Bbb R\to\Bbb R$. Since $f(-n)=0\le x<\vol(A\cap[-n,n])=f(n)$, the regular intermediate value theorem implies that there is some $z$ such that $\vol(A\cap[-n,z])=x$, and for this $z$, $B=A\cap[-n,z]$ satisfies the requirements of the conclusion. $$\tag*{\blacksquare}$$

We need this theorem to extend the formula $\int_A c\,dx=c\vol(A)$ to the case $\vol(A)=\infty$. (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 $\vol(A)=\infty$ and $c>0$, then for any $x\in\Bbb R$ there is a $B\subseteq A$ with $\vol(B)=x/c$, and then $\int_B c\,dx=x\le\int_A c\,dx$. Thus $\int_A c\,dx$ is larger than any real number, so $\int_A c\,dx=\infty$ as well. (This is theorem itg2const2.)

Now we can prove the main theorem.

Proof (of itggt0): Define the sequence $A_n=\{x:f(x)>\frac1n\}$. This is clearly an increasing sequence of measurable sets (since $f$ is measurable), and since $f(x)>0$ for all $x\in A$ so that there is some $k$ with $f(x)>\frac1k$, $A\subseteq\bigcup_{n\in\Bbb N}A_n$. Now assume that $\int_A f(x)\,dx=0$. Then $$0\le\int_{A_n}\frac1n\,dx\le\int_{A_n}f(x)\,dx\le\int_A f(x)\,dx=0,$$ so from itg2const and itg2const2, $\vol(A_n)=0$ (because $\vol(A_n)=\infty$ implies $\int_{A_n}\frac1n\,dx=\infty\ne0$ and $\vol(A_n)\in\Bbb R$ implies $n\int_{A_n}\frac1n\,dx=\vol(A_n)=0$). Since this is true for every $n$, we have $\vol(A)\le\vol(\bigcup_{n\in\Bbb N}A_n)=\lim_{n\to\infty}\vol(A_n)=0$, which contradicts the hypothesis $\vol(A)>0$. Therefore $\int_A f(x)\ne0$, and since $\int_A f(x)\ge0$ we in fact have $\int_A f(x)>0$. $$\tag*{\blacksquare}$$