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Sorry for my bad english.

Let the Dirac measure $\delta_{\alpha}$ for $\alpha \in \mathbb{R}$. Consider $\mu = \sum_{n=1}^{\infty} e^{-n} \delta_{1/n}$ and $\nu = \sum_{n=1}^{\infty} e^n \delta_{1/n}$, two measures on $(\mathbb{R}, B(\mathbb{R}))$.

Firsly, I have to calculate $\mu(\mathbb{R})$ and $\nu(\mathbb{R})$. I've done :

$\mu(\mathbb{R}) = \sum_{n=1}^{\infty} e^{-n} \delta_{1/n}(\mathbb{R}) = \sum_{n=1}^{\infty} e^{-n} = \dfrac{1}{1-e^{-1}}-1 $.

$\nu(\mathbb{R}) = \sum_{n=1}^{\infty} e^{n} \delta_{1/n}(\mathbb{R}) = \sum_{n=1}^{\infty} e^{n} = \dfrac{1}{1-e}-1 $.

Is it right ?

Secondly, I have, for an integer $k \geq 1$, to calculate $\int_{[0,1/k]}d\mu$ and $\int_{[0,1/k]} d\nu$ and determine their limits for $k \rightarrow \infty$. Are they equal to $\mu(\{0\})$ and $\nu(\{0\})$ respectively ?

Despite my lessons, I don't know how to do this second point, at least the beginning. Someone could help me ? Thank you in advance.

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    $\begingroup$ Actually $\sum_{n=1}^{\infty}e^n = + \infty$ ! It's a divergent series. $\endgroup$
    – Crostul
    Apr 24, 2017 at 18:39
  • $\begingroup$ Yes, you're right... ! $\endgroup$ Apr 24, 2017 at 18:47
  • $\begingroup$ $\int_{[a,b]} d \mu = \sum_{n} e^{-n}1_{ 1/n \in [a,b]}$ $\endgroup$
    – reuns
    Apr 24, 2017 at 19:18
  • $\begingroup$ Oh, thank you for the formula ! I didn't know it. $\endgroup$ Apr 24, 2017 at 20:38
  • $\begingroup$ It seems my first answer is also false... :( And $\int_{[0,1/k]} d\mu = \sum_{n \geq 1} e^{-n} 1_{[0,1/k]} (\frac{1}{n}) = \begin{cases} \sum_n e^{-n} & \text{ if } \frac{1}{n} \in [0,1/k] \\ 0 & \text{ if } \frac{1}{n} \notin [0,1/k] \end{cases}$ ? I really need help, I'm blocked.. $\endgroup$ Apr 24, 2017 at 21:19

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So firstly its correct that $\mu(\mathbb R) = \frac{1}{1-e^{-1}}-1$. Secondly as people has said in the notes we have that $\nu (\mathbb R) = \infty$ (because the series is divergent). So now for the integrals:

\begin{align*} \int_{[0,\frac1k]} d\mu &= \mu([0, \frac1k]) = \sum_{n=k}^\infty e^{-n} = \frac{1}{1-e^{-1}} - \sum_{n=0}^{k-1} e^{-n} =\frac{1}{1-e^{-1}}- \frac{1-{e^{-1}}^k}{1-e^{-1}} = \frac{e^{-k}}{1-e^{-1}}.\\ \int_{[0,\frac1k]} d\nu &= \nu([0,\frac1k])=\sum_{n=k}^\infty e^n = \infty. \end{align*}

Hope that was useful.

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