Proposition

If $f: A \rightarrow \mathbb{R}$ is non-negative and $\int_A f = 0$, show that $\{ x : f(x) \ne 0 \}$ has measure 0.

Solution

Let $n \in \mathbb{N}$ be given. We will show that $\{ x : f(x) > 1 / n \}$ has content 0. Let $\epsilon > 0$ be given. Since $\inf \{ U(f, P) \} = 0$, there exists a partition $P$ such that $U(f, P) < \epsilon / n$. Let $S_1, \cdots, S_k$ denote the subrectangles in which $f$ achieves a value higher than $1 / n$. Note that there are only finitely many subrectangles because a partition must only contain finitely many subrectangles.

Since $f$ is non-negative, $M_S \geq 0$ for each subrectangle $S$. Thus $U(f, P) \geq \sum v(S_i) / n$, so $\sum v(S_i) \leq n U(f, P) < \epsilon$.

Therefore, $S_1, \cdots, S_k$ form a finite cover of $\{ x : f(x) > 1 / n \}$ by closed rectangles such that the total volume is less than $\epsilon$. Thus $\{ x : f(x) > 1 / n \}$ has content 0.

Since $n$ was chosen arbitrarily, $\{ x : f(x) > 1 / n \}$ has content 0 for any $n \in \mathbb{N}$. We will use this to show that $\{ x : f(x) \ne 0 \}$ has measure 0. Since $f$ is non-negative, it suffices to show that $\{ x : f(x) > 0 \}$ has measure 0.

Let $\epsilon > 0$ be given.

For each $n \in \mathbb{N}$, let $C_n$ denote a finite cover of $\{ x : f(x) > 1 / n \}$ by closed rectangles such that the total volume is less than $\epsilon / 2^n$. This is always possible because each $\{ x : f(x) > 1 / n \}$ has content 0.

Let $C = \bigcup_{n \in \mathbb{N}} C_n$. Then $C$ covers $\{ x : f(x) > 0 \}$ since for each $f(x) > 0$, there exists an $n \in \mathbb{N}$ such that $f(x) > 1 / n$. Since the countable union of finite sets is at most countable by the diagonal argument on P.50, $C$ is countable. Moreover, $\sum_{S \in C} v(S) = \sum_{i=1}^{n} \sum_{S \in C_i} S < \sum_{i=1}^{n} \epsilon / 2^i = \epsilon$. (Note that the order in which we sum the volumes does not matter. See Theorem 3.55 on P.78 of Principles of Mathematical Analysis.) Thus we found a countable cover of $\{ x : f(x) > 0 \}$ by closed rectangles such that $\sum_{i=1}^{\infty} v(U_i) < \epsilon$.

Therefore, $\{ x : f(x) > 0 \}$ has measure 0.