Integration with respect to counting measure when $X=\mathbb{R}$

Question

Consider the measure space $(X,\mathcal{M},\mu)$ where $X=\mathbb{R}$, $\mathcal{M}=\mathcal{P}(X)$ and $\mu$ is the counting measure. I want to show that for every measurable function $f : \mathbb{R} \mapsto [0,+\infty]$, given $x \in \mathbb{R}$, $\int_{\{ x\}} f \, d \mu = f(x) $. Generally speaking $\int_{X} f \, d \mu = \sum_{x \in X} f(x) $. I can do this in the case $X=\mathbb{N}$ with a monotone convergence argument, but when $X$ is uncountable I find some trouble. How can I fix things when $X$ is uncountable?

Answer 1

First note that by definition,

$$\int_{\{x\}} f\;d\mu=\int f\cdot1_{\{x\}}d\mu.$$

However, note that $f\cdot 1_{\{x\}}=f(x)1_{\{x\}}$. Therefore we use linearity of the integral to compute

$$\int_{\{x\}}f\;d\mu=\int f(x)1_{\{x\}}d\mu=f(x)\int 1_{\{x\}}d\mu=f(x)\,\mu(\{x\})=f(x)$$

where the last equality is by definition of the counting measure.

I'd recommend working on the more general case on your own, using what we just proved.