# $u(j)=1/j^s$ is Lebesgue for the counting measure

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 $$\sum_k f(k)$$