TITLE: Integration with respect to Dirac measure
QUESTION [15 upvotes]: The Dirac measure is defined by $$\delta_x(A)= \begin{cases} 1 &\text{if $x \in A$}\\ 0 &\text{if $x \notin A$}\\ \end{cases}$$ Let $f:X\rightarrow \mathbb{R}$ be a function. Can anyone show me why $\int f \, d\delta_x=f(x)$? Any help is appreciated.

REPLY [16 votes]: Let $g$ denote the constant function being equal to $f(x)$. Then $f=g$ almost surely with respect to $\delta_x$, i.e. the function $f$ is almost surely equal to the constant function taking on the value $f(x)$. This is easily seen since
$$
\delta_x(\{g=f\})=\delta_x(\{y\in X\mid g(y)=f(y)\})=1
$$
since $x\in \{g=f\}$ is in that set. Now use that two functions who are identical almost surely have the same integral. That is
$$
f=g\quad\delta_x\text{-a.s}\;\Longrightarrow\;\int f\,\mathrm d\delta_x=\int g\,\mathrm d\delta_x=f(x)\int 1\,\mathrm d\delta_x=f(x)\cdot\delta_x(X)=f(x).
$$