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

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

Compute

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

Any help is appreciated.

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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 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 at 9:03

1 Answer 1

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 $f_n(x) = x \cdot 1_{K_n}(x)$.

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It is clear. Thank you. –  user73793 Apr 4 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 at 13:15

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