# Prove that $f\in L^{p}$ for $1<p\leq \infty$ with the counting measure.

I'm stuck in this problem from Bartle's Book Elements of Integration. The problem essentially says: Consider the measure space $(\mathbb{N},\mathcal{P}(\mathbb{N}),\mu)$, where $\mu$ es the counting measure. If $f$ is defined on $\mathbb{N}$ by $f(n)=1/n$, then $f\not\in L^{1}$ and $f\in L^{p}$ for $1<p\leq\infty$. Where I'm stuck is when $p=\infty$.

My idea is based on my conjecture that the only element $N\in\mathcal{P}(\mathbb{N})$ such that $\mu(N)=0$ is the empty set. So, from this it follows that $||f||_{\infty}=1$ and I'm done. This is correct?

Bartle defines: the pace $L_{\infty}=L_{\infty}(X,\mathbf{X},\mu)$ consists of all the equivalence classes of $\mathbf{X}$-measurable real-valued functions which are almost everywhere bounded. So if $f\in L^{\infty}$ and $N\in\mathbf{X}$ with $\mu(N)=0$, we define $$S(N)=\sup\{|f(x)|:x\not\in N\}$$ and $$||f||_{\infty}=\inf\{S(N): N\in\mathbf{X}, \mu(N)=0\}$$

So, my idea is correct or not at all? Can you give me any hint?

Thanks.

• Whatever the measure, a function that is everywhere $\le 1$ in absolute value has $L^\infty$ norm no more than $1.$ – zhw. Jul 1 '17 at 16:59

$||f||_\infty=sup_{n \in \mathbb{N}}|f(n)|$. You can think about your $f =(f_1,f_2,\dots)$ as a sequence with values in $\mathbb{R}$. For $p=\infty$ the norm is simply the smallest bound on the absolute value of all elements. Clearly $||f||_\infty=1$ holds in this case.
In addition, if you work with the counting measure $\zeta$, the empty set is the only set with measure $\zeta =0$, all other sets contain at least one element and $\zeta > 0$ clearly.