Integrals and Dirac delta measures Why is it the case that in general $\int X d\delta_k(w) = X(k) $?? 
I can't seem to work it out but it seems like a useful fact? 
 A: Here's two methods of how to show it. Someone correct me if I made some mistakes.
Method 1. It can be proven in steps, starting from simple functions and moving on to measurable functions. (Note that since the whole powerset is the collection of $\delta_{k}$-measurable sets, then every function is $\delta_{k}$-measurable.)
Assume first that $X$ is a simple function, and let $\sum_{i=1}^{n}a_{i}\chi_{A_{i}}$ be its normal representation. Then
\begin{align*}
\int X\,d\delta_{k}=\sum_{i=1}^{n}a_{i}\int \chi_{A_{i}}\,d\delta_{k}=\sum_{i=1}^{n}a_{i}\delta_{k}(A_{i})=\sum_{i=1}^{n}a_{i}\chi_{A_{i}}(k)=X(k),
\end{align*}
since $\delta_{k}(A_{i})=1$ if $k\in A_{i}$ and $0$ otherwise, which is indeed the same as the indicator function of $A_{i}$ at point $k$ (which is indeed the key point to notice).
Assume then that $X$ is non-negative and measurable. There now exists a nondecreasing sequence of simple functions $(\psi_{i})_{i=1}^{\infty}$ so that $\psi_{i}\to X$ pointwise. Using the monotone convergence theorem $(*)$ and the previous step $(**)$ we obtain:
\begin{align*}
\int X\,d\delta_{k} &=\int \lim_{i\to\infty}\psi_{i}\,d\delta_{k}\overset{(*)}{=}\lim_{i\to\infty}\int \psi_{i}\,d\delta_{k}\overset{(**)}{=}\lim_{i\to\infty} \psi_{i}(k)=X(k).
\end{align*}
The result then finally follows by composing any measurable function $X$ to $X=X^{+}-X^{-}$ and invoking the previous step to both of them.
Method 2: Use the fact that for $\delta_{k}$-a.e. we have $X=X(k)$ and integrate.
