In Gerald Folland:
Given a Borel probability measure $\lambda$ on $\mathbb{R}$, one can simply speak of the mean $\bar{\lambda}$ and variance $\sigma^2$ of $\lambda$, $$\bar{\lambda} = \int td\lambda(t), \quad \sigma^2 = \int (t-\bar{\lambda})^2d\lambda(t),$$ which are the mean and variance of any random variable with distribution $\lambda$.
So, is this like the formal definition of the mean (expected value) given not only the standard distribution $P_X$ (given by the CDF $F_X(x)$), but for any measure $\lambda$? Then, in one of the exercises, it says
Let $\delta_t$ denote the point mass at $t \in \mathbb{R}$. Given $a > 0$, let $\lambda_a = e^{-a}\sum_{0}^{\infty}(a^k/k!)\delta_k$ the Poisson distribution with parameter $a$. Show that the mean and variance of $\lambda_a$ are both $a$.
I know how to prove this based on the probability function $p(x;a) = \frac{e^{-a}a^x}{x!}$ (classic sense), but I don't understand the point of view of measure theory
The distribution of a random variable $X$ is the measure $\mu$ on $\mathbb R$ so that for any Borel set $A$: $$ \mu(A) = \Pr(X \in A) .$$ So a a random variable which takes the value $k$ with probability $p_k$ has distribution $$\mu = \sum_k p_k \delta_k .$$ That is, $$ \int f \, d\mu = \sum_k f(k) p_k .$$