($f:R\subset \Bbb R^n\to \Bbb R$, $f\geq 0$, $\int\limits_R f(x)\,dx=0$) $\implies$ ($f=0$ almost everywhere)

Question

I want to prove the following

If $R\subset \Bbb R^m$ is a $m$-dimensional rectangle and $f:R\to \Bbb R$ is a non-negative integrable function with $\displaystyle \int\limits_R f(x)\,dx=0$, then $f$ is zero almost everywhere (i.e. the set $\mathrm{supp}(f)=\{x\in R\,|\, f(x)>0\}$ has zero measure).

Here is my try:

For each $n\in \Bbb N$, define $S_n=\left\{x\in R\,|\, f(x)>\dfrac{1}{n}\right\}$. We see that $\displaystyle\mathrm{supp}(f)=\bigcup_{n\in \Bbb N}S_n$ and, since countable unions of zero measure sets is itself a zero measure set, we are done IF we show that each $S_n$ has zero measure. We have $$0=\int\limits_Rf(x)\,dx\geq \int\limits_{S_n}f(x)\,dx\geq \int\limits_{S_n}\frac{1}{n}\,dx=\frac{1}{n}\int\limits_{S_n}\,dx=\frac{1}{n}\mu(S_n),$$ and, therefore, we must necessarily have $\mu(S_n)=0$.

That is my argument. But after thinking a while, I've thought: "But hey! Who said that such integral $\displaystyle \int\limits_{S_n}\,dx$ does exist?" I mean, what is the negation of the claim

"$A$ has zero measure"?

I was tempted to say: "$A$ has positive measure". But, what if $A$ has not measure at all? (i.e. $\displaystyle \int_A\,dx$ does not exist)

So, what's the proper way to prove this fact, if it is true indeed?

Answer 1

Here is a proof that $S_n$ has measure zero, in the context of multivariable calculus.

Given a bounded function $f:R\to \Bbb R$ on a rectangle $R\subset \Bbb R^m$, we know that the lower and upper integrals $\displaystyle\underline{\int}\limits_Rf(x)\,dx$ and $\displaystyle\overline{\int}\limits_R f(x)\,dx$ always exist and $f$ is integrable if $\displaystyle\underline{\int}\limits_R f(x)\,dx=\displaystyle\overline{\int}\limits_R f(x)\,dx:=\int\limits_Rf(x)\,dx$.

Denote $\chi_{n}:R\to \Bbb R$ the characteristic function of $S_n\subset R$. We have

$$\begin{align*} 0\leq \overline{\int}\limits_{S_n}\,dx&:= \overline{\int}\limits_R\chi_{n}(x)\,dx=n\overline{\int}\limits_{R}\frac{1}{n}\chi_n(x)\,dx\leq n\overline{\int}\limits_{R}f(x)\chi_n(x)\,dx\\ &\leq n\overline{\int}\limits_{R}f(x)\,dx=n\int\limits_{R}f(x)\,dx=0. \end{align*}$$

Therefore, $$\exists \overline{\int}\limits_{S_n}\,dx=0.$$

Let's note now that this implies that $S_n$ has zero measure in the sense of multivariable calculus.

Let $\epsilon>0$ be given. Since $$\overline{\int}\limits_R \chi_n(x)\,dx=\inf \{U(\chi_n;P)\,:\, P \text{ is a partition of } R\},$$ and $U(\chi_n;P)\geq0$, for all $P$, the fact $\displaystyle\overline{\int}\limits_R \chi_n(x)\,dx=0$ implies that, there is some partition $P_0$ of $R$ so that $$U(\chi_n;P_0)=\sum_{A\in P_0}M_{A}(\chi_n)\mathrm{vol}(A)<\epsilon,$$ with $M_A(\chi_n)=\sup\{\chi_n(x)\,:\, x\in A\}$. Define $$B=\{A\in P_0\,:\, A\cap S_n\neq \varnothing\}.$$

Then $B$ is a finite collection of rectangles such that $S_n\subset \displaystyle \bigcup_{A\in B}A$ and $$\sum_{A\in B}\mathrm{vol}(A)=\sum_{A\in P_0}M_A(\chi_n)\mathrm{vol}(A)<\epsilon,$$ and this is exactly the definition of "$S_n$ has zero measure" in multivariable calculus (in this case $S_n$ is bounded, in the general case the definition requires the existence of a countable collection of rectangles such that the series sum of their volumes converge and is less than $\epsilon$).

Answer 2

Well the $S_n$ are measurable subsets, since they are preimages of a measurable function of measurable sets: $S_n=f^{-1}((1/n,\infty])$ (moreover R is measurable as a rectangle).Hence $vol(S_n)=\int_adx$ does exist (it may be infinite though, but you already showed that this cannot hold true here).