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I'm trying to prove that if $f \in L^1$ is such that $\lVert f\psi \rVert_{L^1} = 0$ for all $\psi \in C_c(\mathbb{R})$, then $f = [0] \in L^1$.

Here's my attempt:

Let $E_i$ be such that $$\bigcup_{i=1}^{\infty} E_i = \mathbb{R}$$ and $E_i \cap E_j = \emptyset$ for $i \neq j$. Since integrals are measures,

$$ \lVert f\lVert_{L^1} = \int_{\mathbb{R}} \lvert f\rvert \, d\mu = \sum_{i=1}^{\infty} \int_{E_i} \lvert f\rvert \chi_{E_i} \, d\mu. $$ Let $\phi \in C_c(\mathbb{R})$ be such that $ 0\leq \phi_i \leq 1$ and $\operatorname{supp}(\phi_i) \subset E_i$. Let $\psi \in C_c(\mathbb{R})$ be such that $\infty > \phi_i \geq 1$ and $ E_i \subset \operatorname{supp}(\psi_i).$ Then, by montonicity of the integral $$ 0 = \sum_{i=1}^{\infty} \int_{E_i} \lvert f\phi_i\rvert \, d\mu \leq \sum_{i=1}^{\infty} \int_{E_i} \lvert f\rvert \chi_{E_i} \, d\mu \leq \sum_{i=1}^{\infty} \int_{E_i} \lvert f\psi_i\rvert \, d\mu = 0. $$
It follows that $\lVert f\rVert_{L^1} = 0$.

Is this argument valid? Is there a slicker argument?

Anthony Peter
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    Dynkin's multiplicative system theorem produces a pretty slick argument. – Nate Eldredge Oct 21 '15 at 04:19
  • @Nate Eldredge Indeed it does. :) is the above correct? – Anthony Peter Oct 21 '15 at 04:20
  • Pretty much. You have to use a little bit of care in constructing the sets $E_i$ to ensure that there actually exist $\phi_i, \psi_i$ with the desired properties. And I think you want to demand not only that $E_i$ is in the support of $\psi_i$, but actually that $\psi_i \ge 1$ on $E_i$. And your final "it follows that" needs to invoke Tonelli's theorem, or monotone convergence, or something of the kind. – Nate Eldredge Oct 21 '15 at 04:25
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    By the way, Dynkin's theorem will let you prove a somewhat more difficult fact: if $\int f \phi = 0$ for all $\phi \in C_c(\mathbb{R})$ then $|f|_{L^1} = 0$. This is like your assertion except you are integrating $f$ instead of $|f|$. – Nate Eldredge Oct 21 '15 at 04:26
  • @NateEldredge Could you explain briefly why Tonelli or monotone are needed? – Anthony Peter Oct 21 '15 at 04:30
  • How else do you relate $\sum_i \int |f| \chi_{E_i},d\mu$ to $\int |f|,d\mu$? – Nate Eldredge Oct 21 '15 at 04:31
  • @NateEldredge Ah well I was just using the theorem in Big Rudin that integrals are measures and $|f| \chi_{E_i} = |f|$ on $E_i$. I mean the sum comes from the countable sub-additivity and nothing more – Anthony Peter Oct 21 '15 at 04:33
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    @NateEldredge The proof of that theorem in Rudin uses MCT by the way – Anthony Peter Oct 21 '15 at 04:37

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