Integrating the Poisson Measure Let the set $\mathbb R$ have Borel algebra $\mathcal{X}$comprising of all subsets of $\mathbb R$. Let $\lambda$ be a positive real number. Then the measure $\mu$ is defined as follows: for each $X$ in $\mathcal{X}$
$$
\mu (X) = \sum_{k\,\in\,\mathbb N\,\cap\, X} \frac{e^{-\lambda}\lambda^k}{k!}
$$
Then calculate
$$
\int_{\mathbb R} x\mu (dx)
$$
What have I done so far: well, we are not given that the measure is from Poisson-land (the CDF), I kind of figured that out. I also think the following are true, although it may not be as useful: 


*

*$\mu(\mathbb R\setminus\mathbb N) =0$

*$\mu(\mathbb N) =1$,

*$\mu(\mathbb R) =1$.


What I am not really sure about is should I just substitute the $\mu$ and evaluate the integral (how would that even work). Any hints or suggestions?
 A: Hint: this is just $$e^{-\lambda} \sum\limits_{k=0}^{\infty} k \frac{\lambda^k}{k!}$$
Let me know if you need help computing this. 
A: A discrete measure is one made up entirely of point masses, i.e. there is a subset $\mathbb N$ of $\mathbb R$ with the property that for every $X\in\mathcal X$,
$$
\mu(X) = \sum_{x\,\in\,\mathbb N \, \cap \, X} \mu(\{ x\}). 
$$
The singleton of every member of this subset has positive measure and no set has a measure larger than what that condition necessitates.  With a discrete measure $\mu$, the integral is a sum:
$$
\int_X f(x) \mu(dx) = \sum_{x\,\in\,X\,\cap\,\mathbb N} f(x) \mu(\{x\}). \tag 1
$$
So you can take it to be an exercise that the definition of Lebesgue integration and the definition of sum of a family of nonnegative numbers always yield the same result in $(1)$.
The definition of the sum of a family of nonnegative numbers can be taken to be
$$
\sum_{x\,\in\,\text{some specified set}} f(x) = \sup\left\{ \sum_{x\,\in\, A} f(x) : A\text{ is a finite subset of the specified set.} \right\} \tag 2
$$
Another exercise is to show that $(2)$ agrees with the usual $\varepsilon$-$N$ definition of sum of an infinite series.  One difference between $(2)$ and the usual $\varepsilon$-$N$ definition of sum of an infinite series is that $(2)$ doesn't assume the terms appear in any particular order.  That certainly doesn't matter when we've assumed they're all non-negative.  Another difference is that we have not assumed in advance that the "specified set" is at most countably infinite.  (Another exercise is to show that if there are uncountably many $x$ in the specified set for which $\mu(\{x\})>0$, then the sum is infinite.  For that, just think about how many values of $x$ there are for which $\mu(\{x\})$ is in each of the intervals $\left[\frac 1 {n+1}, \frac 1 n\right)$.)
So the problem reduces to evaluating $\displaystyle \sum_{x=0}^\infty x \frac{\lambda^x e^{-\lambda}}{x!}$.
