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


$\int f d\mu=\sum\limits_{n=1}^{\infty} f(n)$ 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 $\sum_k f(k)$


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