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

Let $\mu:P(\mathbb R) \to [0,+\infty]$ be a measure defined by: $$ \mu (\{ \tfrac 1n \})= \tfrac 1n $$

and $\mu(E)=0$ if $E \cap \{ \tfrac 1n \}_{n \in N_0} =\emptyset$


$$\int_{\mathbb R} x \,d\mu (x)$$

Any help is appreciated.

share|cite|improve this question
you should go through the details to understand better how measures work, but integration with respect to this kind of measure turns in to a weighted sum at the points in the support. – Callus Apr 4 '14 at 8:56
Yes...if a take a simple function than I understand what you mean about the weighted sum...But I don't understand how to explicity compute that integral... – user73793 Apr 4 '14 at 9:03
up vote 1 down vote accepted

Let $K := \{0\} \cup \{\tfrac 1n \mid n \in \mathbb N\}$.

$\displaystyle\qquad \int_{\mathbb R} x \,d\mu = \int_K x\,d\mu + \int_{\mathbb R\setminus K} x \,d\mu = \int_K x\,d\mu$

since $\mu(\mathbb R \setminus K) = 0$.

Now let $K_0 = \{0\}$ and $K_n = K_{n-1} \cup \{\tfrac1n\}$ such that $K = \cup_{n\in\mathbb N_0} K_n$.

Next apply monotone convergence to $f_n(x) = x \cdot 1_{K_n}(x)$.

share|cite|improve this answer
It is clear. Thank you. – user73793 Apr 4 '14 at 9:52
Now if $f_m(x)=x^m$ how can I compute the limit of the integral for $m \rightarrow \infty$?? – user73793 Apr 4 '14 at 13:15

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.