# Thread: Evalute an integral with a weird measure.

1. ## Evalute an integral with a weird measure.

Let $\displaystyle \mu$ be a measure on the set of natural numbers.

And define $\displaystyle \mu ( \{ n, n+1, ... \} ) = \frac {n}{2^n}$ for $\displaystyle n = \{ 1, 2, 3, ... \}$

Find $\displaystyle \int x d \mu (x)$

My solution so far:

To be honest I'm a bit lost in this one. So should I have something like $\displaystyle \sum ^ \infty _{x=1}\frac {x}{2^x }$?

2. ## Re: Evalute an integral with a weird measure.

Let $\displaystyle \mu$ be a measure on the set of natural numbers.

And define $\displaystyle \mu ( \{ n, n+1, ... \} ) = \frac {n}{2^n}$ for $\displaystyle n \in \mathbb {N}$
Taking $\displaystyle n=0$ we get that $\displaystyle \mu(\mathbb N)=0$: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 $\displaystyle \sum_{n=1}^{+\infty}\int_{\{n\}}xd\mu(x)$ (use monotone convergence theorem) and you are almost done. Compute the measure of $\displaystyle \{n\}$ for $\displaystyle n\geq 1$.

5. ## Re: Evalute an integral with a weird measure.

So would I have $\displaystyle \sum ^ \infty _{n=1} x \frac {n}{2^n}$?

6. ## Re: Evalute an integral with a weird measure.

So would I have $\displaystyle \sum ^ \infty _{n=1} x \frac {n}{2^n}$?
There shouldn't be any $\displaystyle x$ in the sum, since you sum over $\displaystyle n$. We have $\displaystyle \mu(\{n\})=\mu(\{n,n+1,\ldots\})-\mu(\{n+1,\ldots\})=\frac{n}{2^n}-\frac{n+1}{2^{n+1}} = \frac{2n-n-1}{2^{n+1}}$.
We have $\displaystyle \int_{\{n\}}xd\mu(x)=n\mu(\{n\})$, since on $\displaystyle \{n\}$ we must have $\displaystyle x=n$.