If $\int f=0$ and $f(x) \ge 0$ for all $x \in \mathbb{R}^d$, then $f=0$ a.e.

Question

If $\int f=0$ and $f(x) \ge 0$ for all $x \in \mathbb{R}^d$, then $f=0$ a.e.

I let $E \subset \mathbb{R}^d$ be a finite measurable set.

I try to break this into two cases:

Case 1: If $f(x)=0$ on $E$, then we are done (as trivially $f=0$ a.e.).

Case 2: If $f(x) > 0$ on $E$, then $\lim_{n \to \infty} \sum_{k=1}^n a_k m(E_k)=0$. I think I can prove by contradiction here. Can I assume $f \not= 0$, which means $f > 0$ in this case? Then, do I need to show that $\int_E f > 0$, which would contradict the hypothesis of $\int_E f = 0$. Is this a good approach to this?

Answer 1

Here's another way to do it which doesn't explicitly make use of approximating $f$ by simple functions.

For $n \in \mathbb{N}$, let

$$E_n = \left\{x \in \mathbb{R}^d \mid f(x) \geq \frac{1}{n}\right\} = f^{-1}\left(\left[\frac{1}{n}, \infty\right)\right).$$

Note that $\displaystyle\bigcup_{n=1}^{\infty}E_n = f^{-1}((0, \infty))$. If $f$ is not zero almost everywhere, then $m(f^{-1}(0, \infty)) > 0$ so there is $n \in \mathbb{N}$ such that $m(E_n) > 0$. Now note that $f \geq \frac{1}{n}\chi_{E_n}$ so $$\int_{\mathbb{R}^d}f\, dm \geq \int_{\mathbb{R}^d}\frac{1}{n}\chi_{E_n}\, dm = \frac{1}{n}m(E_n) > 0.$$

Therefore, if $\displaystyle\int_{\mathbb{R}^d}f\, dm = 0$, $f = 0$ almost everywhere.

Answer 2

Recall that if $f$ is a positive measurable function, there exists an increasing sequence of simple functions $\varphi_n = \sum_{i=1}^{m_n} a_{n,i} \mathbb{1}_{E_{n,i}}$ which converges to $f$, where $\mathbb 1_{E_{n,i}}$ denotes the indicator function and the $E_{n,i}$'s are measurable sets. In the case of $f$ positive, we have $$ \int f \, d\lambda = \sup_{n \in \mathbb N} \int \sum_{i=1}^{m_n} a_{n,i} \mathbb{1}_{E_{n,i}} = \sup_{n \in \mathbb N} \sum_{i=1}^{m_n} a_{n,i} \lambda(E_{n,i}) = 0. $$ In particular, since all the $a_{n,i}$'s are strictly positive, for any $n_0 \in \mathbb N$ we have $$ 0 \le a_{n_0,i} \lambda(E_{n_0,i}) \le \sum_{i=1}^{m_{n_0}} a_{n_0,i} \lambda(E_{n_0,i}) \le \sup_{n \in \mathbb N} \sum_{i=1}^{m_n} a_{n,i} \lambda(E_{n,i}) = 0 \quad \Longrightarrow \quad \lambda(E_{n_0,i}) = 0. $$ It follows that $f \neq 0$ on a set of measure zero, namely on the set $$ \bigcup_{n \in \mathbb N} \bigcup_{i=1}^{m_n} E_{n,i}. $$ (This is a countable union of sets of measure zero, thus has measure zero.)

Hope that helps,