Integral of a product of a ${L^1}_{loc}$ function and a compact support function I'd like to prove that, if $f\in {L^1}_{loc}(\mathbb{R}^n)$ and $g$ is a bounded, measurable function with compact support and if $\int_{\mathbb{R}^n} fgdx = 0$ $\forall g$, then $f$ is almost everywhere equal to $0$.
Surely $fg\in L^1(\mathbb{R}^n)$: by hypothesis, $|g|\leq c\in\mathbb{R}$, so we have
$||fg||_1=\int_{\mathbb{R}^n} |fg| dx\leq c\int_{K}|f|dx\leq +\infty$, where $K$ is the support of the function $g$. So I applied Lebesgue differentiation theorem to $fg$, in order to have
$lim_{r\to 0^{+}} \frac{1}{|B(x,r)|} \int_{B(x,r)} f(y)g(y)dy = f(x)g(x)$ almost everywhere, but now I don't know how to use the hypothesis about the integral of this product over $\mathbb{R}^{n}$. How can I proceed?
 A: Let $g$ be bounded, measurable and have compact support. Define $A := f^{-1} ([0,\infty)) \in \mathcal B (\Bbb R^n)$. Then $g\cdot 1_{A}$ is a measurable, bounded function with compact support. Hence
$$0=\int_{\Bbb R^n} f g 1_{A} = \int_{\Bbb R^n} f ^+ g ,$$
where $f^+$ is the positive part of $f$. Thus, without loss of generality we can assume that $f\geq 0$. Now let $g = 1_{B(0,r)}$ with $r>0$. The function $g$ is bounded, measurable and has compact support. Therefore,
$$\int_{B(0,r)} f =0.$$
Since $f\geq 0$, this yields $f = 0$ almost everywhere on $B(0,r)$. This means there exists a null set $N_r \subseteq B(0,r)$ with
$$f(x) = 0 \quad \forall x\in B(0,r) \setminus N_r .$$
This provides
$$f(x) = 0 \quad \forall x \in \bigcup\limits_{m=1}^\infty ( B(0,m) \setminus N_m) \supseteq \Bbb R^n \setminus \bigcup_{m=1}^\infty N_m. $$
Since $\lambda (\bigcup_{m=1}^\infty N_m) \leq \sum_{m=1}^\infty \lambda (N_m) = 0$ we have that $f=0$ almost everywhere.
A: Since $g = 1_{B(x,r)}$ is measurable and bounded with compact support you could also proceed with your idea as follows:
$$\int_{B(x,r)} f =  \int_{\Bbb R^n} fg = 0$$
Thus for almost every $x\in\Bbb R^n$ $$0 = \lim_{r\searrow 0} \frac{1}{\vert B(x,r)\vert} \int_{B(x,r)} f  = f(x)$$
