text: # Thread: Evalute an integral with a weird measure.
  
  1. ## Evalute an integral with a weird measure.
  
  Let $\mu$ be a measure on the set of natural numbers.
  
  And define $\mu ( \{ n, n+1, ... \} ) = \frac {n}{2^n}$ for $n = \{ 1, 2, 3, ... \}$
  
  Find $\int x d \mu (x)$
  
  My solution so far:
  
  To be honest I'm a bit lost in this one. So should I have something like $\sum ^ \infty _{x=1}\frac {x}{2^x }$?
  
  2. ## Re: Evalute an integral with a weird measure.
  
  Let $\mu$ be a measure on the set of natural numbers.
  
  And define $\mu ( \{ n, n+1, ... \} ) = \frac {n}{2^n}$ for $n \in \mathbb {N}$
  Taking $n=0$ we get that $\mu(\mathbb N)=0$:maybe there is a typo.
  
  3. ## Re: Evalute an integral with a weird measure.
  
  Sorry, n should be equal to 1, 2, 3, ... I changed it in the original post.
  
  4. ## Re: Evalute an integral with a weird measure.
  
  Ok. Hint: write the integral as $\sum_{n=1}^{+\infty}\int_{\{n\}}xd\mu(x)$ (use monotone convergence theorem) and you are almost done. Compute the measure of $\{n\}$ for $n\geq 1$.
  
  5. ## Re: Evalute an integral with a weird measure.
  
  So would I have $\sum ^ \infty _{n=1} x \frac {n}{2^n}$?
  
  6. ## Re: Evalute an integral with a weird measure.
  
  So would I have $\sum ^ \infty _{n=1} x \frac {n}{2^n}$?
  There shouldn't be any $x$ in the sum, since you sum over $n$. We have $\mu(\{n\})=\mu(\{n,n+1,\ldots\})-\mu(\{n+1,\ldots\})=\frac{n}{2^n}-\frac{n+1}{2^{n+1}} = \frac{2n-n-1}{2^{n+1}}$.
  We have $\int_{\{n\}}xd\mu(x)=n\mu(\{n\})$, since on $\{n\}$ we must have $x=n$.
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