# Calculate $\int_{[0,1/k]} d\mu$ with $\mu$ measure

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.

• Actually $\sum_{n=1}^{\infty}e^n = + \infty$ ! It's a divergent series. Apr 24, 2017 at 18:39
• Yes, you're right... ! Apr 24, 2017 at 18:47
• $\int_{[a,b]} d \mu = \sum_{n} e^{-n}1_{ 1/n \in [a,b]}$ Apr 24, 2017 at 19:18
• Oh, thank you for the formula ! I didn't know it. Apr 24, 2017 at 20:38
• 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.. Apr 24, 2017 at 21:19

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: