I was wondering whether the following statement is true and, if so, how it can be shown:
If $ f \in L^{1}_{Loc}(\mathbb{R}^n) $ and if for all compactly supported continuous functions $ g: \mathbb{R}^n \to \mathbb{C} $ we have that the Lebesgue integral of $ f $ multiplied by $ g $ equals zero, i.e. $$ \int_{\mathbb{R}^n} f(x)g(x) \mathrm{d}x = 0 , $$ then $ f(x) = 0 $ almost everywhere.
I would be very grateful for any answers or hints!
N.B. I am aware that this question has already been addressed. In If $f\in L^1_{loc}(\mathbb{R})$ and $\int f\varphi=0$ for all $\varphi$ continuous with compact support, then $f=0$ a.e., I am not quite sure about how to create a sequence of compactly supported continuous functions such that $ \varphi_n\to \frac{f}{|f|+1} $. This particular question may have its answer in If $f\in L^1(\mathbb{R})$ is such that $\int_{\mathbb{R}}f\phi=0$ for all continuous compactly supported $\phi$, then $f\equiv 0$., however here I am unsure about the meaning of a "regularizing sequence"; why does $ \phi_n\ast f\to f $ in $L^1$ sense if $ \phi_n(x) = n\phi(nx) $, where $ \phi\in \mathcal C^\infty_c(\Bbb R) $ with $ \phi\ge 0 $ and $ \int_{\Bbb R}\phi(x)dx=1 $?
Once again, any answer would be much appreciated!
Best Answer
For any ball $B(a,r),$ there is a sequence $g_k$ of continuous functions with support in $B(a,r),$ with $|g_k|\le 1$ everywhere, such that $g_k(x) \to \text { sgn }(f(x))$ pointwise a.e. in $B(a,r).$ Thus
$$\tag 1 \int_{B(a,r)} |f| = \int_{B(a,r)} f\cdot \text { sgn }(f) = \lim \int_{B(a,r)} f\cdot g_k =\lim 0 =0.$$
The second equality in $(1)$ follows from the dominated convergence theorem. From $(1)$ we see $f=0$ a.e. on $B(a,r).$ Since $B(a,r)$ was an arbitrary ball, we have $f=0$ a.e. in $\mathbb R^n.$