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Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$ and $f\in\mathcal L^1_{\text{loc}}(\lambda)$ with $$\int\varphi f\:{\rm d}\lambda=0\;\;\;\text{for all }\varphi\in C_c(\mathbb R)\tag1.$$

I want to conclude that $\lambda$-a.e. $f=0$.

Let $\mu$ denote the measure with density $|f|$ with respect to $\lambda$. By assumption $$\mu(K)<\infty\;\;\;\text{for all compact }K\subseteq\mathbb R\tag2$$ and hence we know that $C_c(\mathbb R)$ is dense in $L^1(\mu)$. Now, let $(K_n)_{n\in\mathbb N}$ be a nondecreasing of compact subsets of $\mathbb R$ with $\bigcup_{n\in\mathbb N}K_n=\mathbb R$. Obviously, it's sufficent to conclude $\mu(K_n)=0$ for all $n\in\mathbb N$.

How can we show that?

We may observe that $$\int\varphi\operatorname{sgn}f\:{\rm d}\mu=\int\varphi f\:{\rm d}\lambda\;\;\;\text{for all }\varphi\in C_c(\mathbb R)\tag3.$$ Fix $n\in\mathbb N$. Clearly, $1_{K_n}\operatorname{sgn}f\in\mathcal L^1(\mu)$ and hence there is a $(\varphi_m)_{m\in\mathbb N}\subseteq C_c(\mathbb R)$ with $$\left\|\varphi_m-1_{K_n}\operatorname{sgn}f\right\|_{L^1(\mu)}\xrightarrow{m\to\infty}0\tag4.$$ The idea should be now to multiplicate $\varphi_m-1_{K_n}\operatorname{sgn}f$ by $\operatorname{sgn}f$ and use $(3)$. However, is this legitimate? And we only have $\operatorname{x}\operatorname{x}=1$ if $x\ne 0$, which could be problematic ...

0xbadf00d
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  • You might want to see https://math.stackexchange.com/questions/2416568/if-int-fg-0-for-all-compactly-supported-continuous-g-then-f-0-a-e?rq=1 – Byeong-Ho Bahn Jan 26 '19 at 00:32
  • @LevBan Thank you, but actually I'm aware of other proofs of the claim. I would like to know how we need to complete the argument in the question. – 0xbadf00d Jan 26 '19 at 00:39
  • @0xbadf00d Your proof easily works. Instead of showing $\mu(K_n) = 0$ for each $n$, you have shown $\mu(K_n\setminus{f = 0}) = 0$ for each $n$, which is obviously sufficient as well. – mathworker21 Jan 29 '19 at 14:46

1 Answers1

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All you need is the following lemma: If $g_n\to g$ in $L^1(\mu),$ and $b$ is bounded and measurable on $\mathbb R,$ then $bg_n\to bg$ in $L^1(\mu).$

Proof: Suppose $|b|\le M$ everwhere. Then

$$\int_{\mathbb R} |bg_n-bg|\, d \mu = \int_{\mathbb R} |b| |g_n-g|\, d \mu \le M\int_{\mathbb R} |g_n-g|\, d \mu \to 0.$$

In your problem we have $\varphi_n \to 1_K\operatorname{sgn}f$ in $L^1(\mu)$ and $b= \operatorname{sgn}f.$ By the lemma,

$$(\operatorname{sgn}f)\varphi_n \to \operatorname{sgn}f\cdot(1_K\operatorname{sgn}f) = 1_K\cdot 1_{\{f\ne 0\}}\,\, \text {in } L^1(\mu).$$

Using your $(3),$ we thus obtain

$$0 = \int_{K\cap \{f\ne 0\}}|f|\,d\lambda = \int_{K}|f|\,d\lambda.$$

Thus $f=0$ $\lambda$-a.e. on $K.$

zhw.
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