Proving $\lim_{p\to 0} ||f||_p^p = \mu(\operatorname{supp}f)$

Question

Q) Let $\mu$ be a counting measure on $\mathbb{N}$. Prove that

$$\lim_{p\to 0} ||f||_p^p = \mu(\operatorname{supp} f)$$

My attempt: I can prove this if the support of $f$ is a finite set. Because then,

$$\lim_{p\to 0} \sum_n |f(n)|^p = \sum_n \lim_{p\to 0} |f(n)|^p = \sum_{n:f(n)\neq 0} 1 = \mu(\operatorname{supp} f)$$

But if $\operatorname{supp} f$ is an infinite set, may I know how to show that $\lim_{p\to 0} ||f||_p^p = \infty$?

Answer 1

Suppose the support if $f$ is an infinite set, say $\{n_1,n_2,...\}$. Then $\sum |f(n)|^{p} \geq \sum\limits_{k=1}^{N} |f(n_k)|^{p} \to N$ as $p \to 0$ for each positive integer $N$. Hence the limit is $\infty$, as required.

Answer 2

You can assume that $|f(n)|<1$ for all $n$ because if there are infinitely many integers for which $|f(n)|\ge 1$ the result is clear. Now, define $f_m(n)=|f(n)|^{1/m}.$ Then $f_m$ is positive increasing so the Monotone Convergence Theorem applies to show that $\underset{m\to\infty}\lim \sum_n |f(n)|^{1/m}=\sum_n \underset{m\to\infty}\lim|f(n)|^{1/m}=\infty.$