Suppose $F$ is given by
$$ F(x) = \sum_{i=1}^{\infty} \frac{1}{2^i} \chi_{[1/i, \infty)}(x) $$
and put $P( ( - \infty , x]) = F(x)$. I am trying to compute $P( [ 1 , \infty)) $, $P( [ \frac{1}{10}, \infty)) $, $P( \{0 \} )$, $P( (0, \infty) ) $. For the first one, I have
$$ P( [1, \infty)) = 1 - P( (-\infty,1)) = 1 - \lim_{x \to 1^{-}} F(x) = 1 -1 =0 $$
$$ P( [ \frac{1}{10}, \infty )) =1 - \lim_{x \to 1/10 ^{-}} F= 1-1 =0 $$
$$ P( (0 , \infty)) = 1 - P( ( -\infty, 0]) = 1 - F(0) = 1 -0 = 1$$
Are these correct? Also, I am stuck trying to find the probability of the singleton $\{0 \}$. Any help would be appreciated.
Note that: $$\chi_{\left[i^{-1},\infty\right)}\left(x\right)=\chi_{\left(-\infty,x\right]}\left(i^{-1}\right)=\delta_{i^{-1}}\left(\left(-\infty,x\right]\right)$$ so that: $$P\left(\left(-\infty,x\right]\right)=F\left(x\right)=\sum_{i=1}^{\infty}2^{-i}\delta_{i^{-1}}\left(\left(-\infty,x\right]\right)$$
In fact we have: $$P=\sum_{i=1}^{\infty}2^{-i}\delta_{i^{-1}}$$
Here $\delta_{x}$ denotes the Dirac measure at $x$.
This spreads light. E.g. it is immediately clear that $P(\{0\})=0$