Stein 2.11

Prove that if $f\in{L^1(\mathbb{R}^d)}$, real-valued, and $\int_E f(x)dx \geq 0$ for every $E\in{\mathcal{M}}$ then $f(x) \geq 0$ a.e. $x$. Similarly, if $\int_E f(x)dx=0$ for every $E\in{\mathcal{M}}$ then $f(x) = 0$ a.e.

$\textit{Proof}$ : This problem can be tackled in various ways using the tools so far. One way is as follows:

Define $A=\{x \mid f(x) < 0 \}$. $f$ is measurable so $ \{f < a \} \in{\mathcal{M}} $, $\forall a\in{\mathbb{R}}$. Letting $a=0$ makes the set $A\in{\mathcal{M}}$. By assumption, $$\int_Af(x)dx \geq 0$$ By the way $A$ is defined, $$f\chi_A(x) \leq 0$$ $$\int_Af(x)dx = \int_\mathbb{R}f(x) \chi_A(x)dx \leq \int_\mathbb{R}0=0$$ Therefore, $$ \int_Af(x)dx=0$$ We have $f(x) < 0$ $\forall x\in{A}$ and $\int_Af(x)dx=0$. Now assume that to the contrary that $m(A) > 0$. Let $Q_n $ be the compact cube centered at the origin with side lengths $n$. So, $A \cap Q_n$ is measurable and bounded and $A\cap Q_n \nearrow_{n\rightarrow \infty} A$. [Then, by corollary 3.3, $m(A)=\lim_{n\rightarrow \infty}m(A \cap Q_n)$.] On $A \cap Q_n$ we have conditions for Lusin's theorem met. So, on a contained closed set $F_\epsilon \subseteq A \cap Q_n$ $f\in{C(F_\epsilon)}$. Additionally, $F_\epsilon$ has positive measure. With $f(x_0) <0$ for some $x_0$ on a set of positive measure, then the integral of $f$ around a neighborhood of $x_0$ must be negative since $f$ is continuous...a contradiction.
$\therefore \quad m(A)=0$ i.e. $f \geq 0$ a.e.

Alternatively, a very quick and direct way of proof is to use one of the main results of chapter 3: the Lebesgue differentiation theorem. This states that if $f$ is integrable (or even just locally integrable) then $$\lim_{x\in{B},m(B)\rightarrow 0} \frac{1}{m(B)} \int_B f(t)dt = f(x) \quad a.e.$$ Balls in $\mathbb{R}^d$ are measurable, so over any ball, $\int_B f(t)dt \geq 0 $. Since $f$ is integrable, we apply the Lebesgue differentiation theorem. So, $$f(x) = \lim_{x\in{B},m(B)\rightarrow 0} \frac{1}{m(B)} \int_B f(t)dt \geq 0 \quad a.e.$$ i.e. $f \geq 0$ a.e. Done.

Similarly, if $\int_E f(x)dx=0$, then we get $$f(x) = \lim_{x\in{B},m(B)\rightarrow 0} \frac{1}{m(B)} \int_B f(t)dt=0 \quad a.e. \quad \quad \blacksquare$$