Problem : $\mu$ and $\nu$ are functions defined on $\mathcal{Bor(R)}$ where $\delta_a$ is the measure of Dirac on $a \in R$
1) Verify that $\mu = \sum^{+\infty}_{n=1} e^{-n} \delta_{1/n}$ and $\nu = \sum^{+\infty}_{n=1} e^{n} \delta_{1/n} $ are measures and define their negligeable sets.
2) Are these measures finite? of probability? are they $\sigma$ finite ?
3) For $m=\mu$ or $\nu$, calculate $m([0;1/k])$ with $k\geq1$ and $lim_{k \rightarrow \infty} m([0;1/k])$ compare the results
Solution:
1) We have that $\delta_1/n(\emptyset)=0$ for all $n\geq1$ so the series associated to $\mu(\emptyset)$ and $\nu(\emptyset)$ converge and $\mu(\emptyset) = 0 = \nu(\emptyset)$. Let $(A_n)_{n\geq1}$ a sequence of disjoint subsets of borelians of $\mathbb{R}$. Since $ \delta_1/n$ is a measure for all integer $k\geq1$ we have :
Where the permutation of the summation is ligit because the double indexed sequence $(\delta_{1/k}(A_n))_{n,k}$ is of positif terms. We obtain the same way :
$\mu$ and $\nu$ are $\mathcal{Bor(R)}$ measure which negligeable sets are :
2) We have that :
and
for all $n \geq 1$
So $\mu$ is finite, so it is $\sigma$-finite but not of probability, and $\nu$ is infinite so not of probability.
Moreover :
and
so equally, $ \nu $ is sigma finite.
3) For all $k \geq 1$ we have : 
(remainder of a convergent series)
But also :
The measure $\mu$ is right continuous at $0$ but not the measure $\nu$ because it is not finite in the neighberhood of $0$.
End of solution
In solution of question 2, isn't $\mu(R)$ supposed to be infinity why are we using $\mu(R)$?
Why are we using the interval $([0;1])$, and why aren't we using the same argument of measure$([0;1])$ for both $\mu$ and $\nu$? Why the interval $[0;1]$ for $\mu$ and $[1/n;1]$ for $\nu$
About the logic of question 2 , can a measure be finite but not $\sigma$ finite?