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 $x=n$.