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Let $\mu$ and $\nu$ be $\sigma$-finite measures on $(\mathbb{R}^k,\mathcal{B}(\mathbb{R}^k))$. If $\int f d\mu = \int f d\nu$ for all continuous functions $f:\mathbb{R}^k\to\mathbb{R}$, then $\mu = \nu$.

This is a homework question, so general strategy and hints for proof will be awesome for me. You can write the exact answer tomorrow, though :)

Thanks!

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  • $\begingroup$ Are the measures finite on compact sets? $\endgroup$ – copper.hat Oct 22 '13 at 23:44
  • $\begingroup$ copper.hat, the above written thing is all of the information given in the question. $\endgroup$ – user48547 Oct 22 '13 at 23:48
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    $\begingroup$ Maybe this question is of use. $\endgroup$ – Stefan Hansen Oct 23 '13 at 6:22
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I suspect that the result is not true.

I suspect that there is an implied (or missing) hypothesis that the measures are bounded on bounded sets.

Let $\mu A = \int_{A \cap (0,1]} \frac{dt}{t}$, and $\nu A = \mu A + \delta_0 A$, where $\delta_0$ is the Dirac measure with point mass at $x=0$.

Clearly $\mu \ne \nu$.

However, if $f$ is continuous, then either $f(0) = 0 $ or not. If $f(0) = 0$, then $\int f d \mu = \int f d \nu$, and if $f(0) \neq 0$, then both integrals are infinite with the same sign. Hence $\int f d \mu = \int f d \nu$ for all continuous $f$.

Both measures are finite on each element of the collection $\{ (-\infty, 0] \} \cup \{(\frac{1}{n+1}, \frac{1}{n}] \}_{n} \cup \{ (1,\infty) \} $.

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  • $\begingroup$ thanks. I will discuss the question with professor. $\endgroup$ – user48547 Oct 23 '13 at 0:17
  • $\begingroup$ There is a subtlety here that leaves me a little uncertain. Stating that the integrals are equal for all continuous $f$ implies that the integrals exist (allowing unambiguous infinite values as well). This would have implications for the measures $\mu,\nu$. $\endgroup$ – copper.hat Oct 23 '13 at 0:18
  • $\begingroup$ Can you tell more about "if f(0)=0, then $\int{f}d\mu=\int{f}d\nu$"? $\endgroup$ – Mike Brown Apr 8 '15 at 5:17
  • $\begingroup$ You have $\int f d \nu = \int f d \mu + f(0)$ for all integrable $f$. $\endgroup$ – copper.hat Apr 8 '15 at 6:22
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    $\begingroup$ @Ben: You have $\int f d \mu = \int_{(0,1]} f(t)\frac{dt}{t}$. $\endgroup$ – copper.hat Apr 9 '15 at 13:40

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