0
$\begingroup$

I am trying to solve the following.

Consider measure space $$(\mathbb{N},\mathcal{P}(\mathbb{N}),\mu)$$ with $\mu$ as the counting measure.

With $s<0$, let $u:\mathbb{N} \rightarrow \mathbb{R}$ as $u(j)=1/j^s$

For which $s>0$ is $u \in \mathcal{L}^1$?

My idea is:

We have the integral $$\int_{\mathbb{N}} (1/j)^s \, d\mu=\sum_{j=1}^\infty (1/j)^s$$

And so we get a p-series which only converges with $s>1$

So that $u\in \mathcal{L}^1$ when $s>1$

However I am in doubt if I can just change the integral to a sum like this because it is on the singletons of natural numbers

Any help would be appreciated

$\endgroup$
2
$\begingroup$

$\int f d\mu=\sum\limits_{n=1}^{\infty} f(n)$ is true for any non-negative function on $\mathbb N$.

$f_n(k)=f(k)$ for $ k \leq n$ and $0$ for $k>n$ defines a sequence of simple functions increasing to $f$ so $\int f d\mu=\lim \int f_n d\mu$ and $(\int f_n d\mu)$ is nothing but the partial sum sequence of the series $\sum_k f(k)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.