Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
This is the defining property of the Dirac delta. – Qiaochu Yuan May 31 '12 at 20:32
Should the right-hand side be $X(0)$? Or at what point is that dirac measure standing on? – T. Eskin May 31 '12 at 20:40
Yes, thankyou I have changed it and I think it is correct now? Can anybody explain why this is the case or is it just by definition? – Rosie May 31 '12 at 20:42
Yeah, it looks correct now. – T. Eskin May 31 '12 at 21:18
up vote 2 down vote accepted

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.

share|cite|improve this answer
Thanks very much that's really helpful, I think method 1 is more in line with my course, could you just explain to me though you have written "The result then finally follows by composing any measurable function X to X=X + −X − and invoking the previous step to both of them". Does this mean that both X+ and X- are integrable? Im a little confused – Rosie May 31 '12 at 21:42
@rosie: We can always do such a composition $X=X^{+}-X^{-}$ where $X^{+}=\max\{X,0\}$ and $X^{-}=-\min\{X,0\}$. Then by using the previous step (since both $X^{+}$ and $X^{-}$ are non-negative and measurable) we get $\int X\,d\delta_{k}=\int X^{+}\,d\delta_{k}-\int X^{-}\,d\delta_{k}=X^{+}(k)-X^{-}(k)=X(k)$. – T. Eskin May 31 '12 at 21:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.