Integration of non-negative function over Dirac's measure

Question

Let $\delta_{x_0}$ be de Dirac delta function, i.e. $\delta_{x_0}: \mathcal{A} \longrightarrow \mathbb{R}$ where $\mathcal{A}$ is a $\sigma$-algebra, and $$\delta_{x_0}(B)=1 \: \mbox{if}\: x_0 \in B \quad ; \quad \delta_{x_0}(B)=0 \: \mbox{if}\: x_0 \notin B$$ I want to prove that if $f: \mathbb{R} \longrightarrow[0, +\infty)$ then $$\int_{\mathbb{R}}fd\delta_{x_0}=f(x_0)$$ The definition I've been given of $\int fd\mu$, where $\mu$ is any measure is this: $$\int_X fd\mu=\sup\left\{ \int_X s(x)d\mu:0\leq s(x) \leq f(x), \: s\mbox{ simple function} \right\}$$ (I haven't learned yet to define the integral of any function; just simple functions and non-negative ones). I don't know how can I conclude that $\int_{\mathbb{R}}fd\delta_{x_0}=f(x_0)$ if I don't know the definition of $f$. Can someone help me?

Answer 1

The key to proving properties of integrals of general positive functions $f$, is that any positive function $f$ can be written as the limit $f=\lim_{n\rightarrow \infty}s_n$, where $(s_n)_{n\in \mathbb{N}}$ is an increasing sequence of simple functions. And we then use the monotone convergence theorem to conclude, that $$\int_X f \: d\mu = \lim_{n \rightarrow \infty} \int_X s_n \: d\mu.$$ Applying this method to the dirac measure $\delta_{x_0}$, we get that \begin{align*}\int_X f(x) \: \delta_{x_0}(dx) &= \lim_{n\rightarrow \infty} \int_X s_n(x) \: \delta_{x_0}(dx) \\ &= \lim_{n\rightarrow \infty} s_n(x_0) \\ &= f(x_0) \end{align*} for any increasing sequence of simple functions $(s_n)_{n\in \mathbb{N}}$ converging to $f$. I leave it for you to verify that $$\int_X s(x) \delta_{x_0}(dx) = s(x_0),$$ when $s$ is simple.