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.

  • $\begingroup$ 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
  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 $f_N:= f\mathbb 1_{ B\cap [0,N]^n}$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.