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Assume $f$ is an integrable function on $\mathbb{R^n}$. Assume for every bounded continuous function g on $\mathbb{R^n}$, $\int_\mathbb{R^n}fg=0$. Prove $f$ must equal $0$ almost everywhere.

I am really not sure how to do this problem. I have tried simple stuff but that is not currently working.

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  • $\begingroup$ hint : This is a simple aplication of the classical Du Bois - Raymond lemma =) . $\endgroup$ – math student Jul 26 '13 at 15:16
  • $\begingroup$ @LeandroTavares Any ideas other than that? I am studying for a comp exam and this is part of a test bank provided to us. However, we did not discuss this lemma. $\endgroup$ – Leo Spencer Jul 26 '13 at 15:23
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  1. Since the characteristic function of a closed set is a pointwise non-increasing limit of continuous functions, we can prove by a monotone convergence argument that $\int_{\mathbb R^n}f\chi_F=0$ for each closed subset $F$ of $\mathbb R^n$.

  2. For each $\varepsilon>0$ and Borel set $S$ of finite measure, we can find a closed subset $F$ contained in $S$ such that $\lambda_n(S\setminus F)<\varepsilon$. Hence $\int_{\mathbb R^n}f\chi_S=0$ for each Borel subset of finite measure.

  3. This can be extended to each Borel subset. Indeed, if $B$ is a Borel subset of $\mathbb R^n$, then $B\cap [0,N]^n$ is a Borel subset of finite measure. By 2., this implies that $\int_{\mathbb R^n } f\mathbb 1_{ B\cap [0,N]^n}=0$ for each $N$. Then use the dominated convergence theorem for the sequence $\left(f_N\right)_N$ defined by $f_N:= f\mathbb 1_{ B\cap [0,N]^n}$.

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  • $\begingroup$ Could you elaborate a bit on how to extend to non-finite measure Borel subset ? $\endgroup$ – Hua Dec 7 '16 at 12:04
  • $\begingroup$ @Hua I haved edited in order to add more details. $\endgroup$ – Davide Giraudo Dec 7 '16 at 12:49
  • $\begingroup$ Oh, I guess I get it. Thanks ! $\endgroup$ – Hua Dec 7 '16 at 12:53
  • $\begingroup$ Sorry but I wish you could clarify me for another point about the step 2. As far as I know, the "approximation" property used in 2 is doable only in so-called inner regular measure, so this proof works under the assumption that the measure in this problem is inner regular, right ? Thanks. $\endgroup$ – Hua Dec 8 '16 at 3:04
  • $\begingroup$ Indeed, it works in this more general context. $\endgroup$ – Davide Giraudo Dec 8 '16 at 14:36

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