$\mathscr{L}^{p_1}(\mathbb{N},\mathscr{A},\mu)\subseteq\mathscr{L}^{p_2}(\mathbb{N},\mathscr{A},\mu)$: $1\leq p_1

Question

I need to prove the following:

Suppose that $1\leq p_1<p_2<+\infty$. Let $\mu$ is the counting measure on the $\sigma$-algebra $\mathscr{A}$ of all subsets of $\mathbb{N}$. Then $\mathscr{L}^{p_1}(\mathbb{N},\mathscr{A},\mu)\subseteq\mathscr{L}^{p_2}(\mathbb{N},\mathscr{A},\mu)$.

My Question:

1. I tried it myself, but I got stuck on a step of my attempt (see below). I am not sure if my existing steps are correct or not either. I would really appreciate it if someone could check my existing work and help me out with where I got stuck.

2. Moreover, I would like to see if there is any easier solution. Thanks a lot for any help!

My Attempt:

Let $f\in\mathscr{L}^{p_1}(\mathbb{N},\mathscr{A},\mu)$. Then $f$ is an $\mathscr{A}$-measurable function on $\mathbb{N}$ such that $|f|^{p_1}$ is integrable. So $\int\left(|f|^{p_1}\right)^+d\mu = \int|f|^{p_1}d\mu < +\infty$. Write $f=\sum_{i=1}^{\infty}a_i\chi_{\{i\}}$. Then $|f|^{p_1} = \sum_{i=1}^{\infty}|a_i|^{p_1}\chi_{\{i\}}$. I want to prove that \begin{align*} \int|f|^{p_1}d\mu &= \sup\left\{\int gd\mu:g\in\mathscr{S}_+\ \text{and}\ g\leq|f|^{p_1}\right\}\\ &= \sum_{i=1}^{\infty}|a_i|^{p_1}\quad (< +\infty). \end{align*}

I couldn't figure out how to prove that $\sum_{i=1}^{\infty}|a_i|^{p_1}$ is the least upper bound of the set $\left\{\int gd\mu:g\in\mathscr{S}_+\ \text{and}\ g\leq|f|^{p_1}\right\}$.

What remains is to apply the following result (see also this post):

Suppose that $1\leq p_1<p_2<=\infty$. Then each sequence $\{a_n\}$ that satisfies $\sum|a_n|^{p_1}<+\infty$ also satisfies $\sum|a_n|^{p_2}<+\infty$

If the part where I got stuck is true, then this result would imply that $\int|f|^{p_2}d\mu<+\infty$, and thus $f\in\mathscr{L}^{p_2}(X,\mathscr{A},\mu)$.

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Note: $\mathscr{S}_+$ is the set of simple nonnegative real-valued $\mathscr{A}$-measurable functoin.

Note: I aware that related questions have been asked here and here. But this question is asking about different stuff.

Thank you very much in advance!

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Update:

Thanks to @ThànhNguyễn's comment. I wrote an answer for this post. I would really appreciate if someone could help me check if it is correct or not! Thank you very much!

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Reference:

Example 3.3.5 from Measure Theory by Donald Cohn.

Answer 1

Thanks to @ThànhNguyễn's comment and help!

Proposition$\quad$ Let $(\mathbb{N},\mathscr{A},\mu)$ be a measure space, where $\mathscr{A}$ is the $\sigma$-algebra of all subsets of $\mathbb{N}$ and $\mu$ is the counting measure on $\mathscr{A}$. Let $f$ be a real- or complex-valued $\mathscr{A}$-measurable function on $\mathbb{N}$. Let $1\leq p<+\infty$. Show that \begin{align*} \int|f|^pd\mu = \sum_{i=1}^{\infty}|f(i)|^p \end{align*}

Proof$\quad$ Since $f$ is a real- or complex-valued $\mathscr{A}$-measurable function on $\mathbb{N}$, $|f|^p$ is a nonnegative real-valued $\mathscr{A}$-measurable function on $\mathbb{N}$. Write $f=\sum_{i=1}^{\infty}f(i)\chi_{\{i\}}$, then $|f|^p = \sum_{i=1}^{\infty}|f(i)|^p\chi_{\{i\}}$ Define a sequence $\{f_n\}$ of simple nonnegative real-valued $\mathscr{A}$-measurable functions on $\mathbb{N}$ by letting \begin{align*} f_n(x) = \sum_{i=1}^n|f(i)|^p\chi_{\{i\}} = \begin{cases} |f(i)|^p\quad &\text{if $x=i$},\\ \\ 0\quad &\text{otherwise}. \end{cases} \end{align*} Then the sequence $\{f_n\}$ satisfies \begin{align*} f_1(x)\leq f_2(x)\leq\dots \end{align*} and \begin{align*} |f(x)|^p = \lim_{n\to\infty}f_n(x) \end{align*} at each $x$ in $\mathbb{N}$. Note that for each $n$, \begin{align*} \int f_nd\mu = \sum_{i=1}^n|f(i)|^p\mu(\{i\}) = \sum_{i=1}^n|f(i)|^p. \end{align*} Then by the Monotone Convergence Theorem, we have \begin{align*} \int|f|^pd\mu &= \lim_{n\to\infty}\int f_nd\mu\\ &= \lim_{n\to\infty}\sum_{i=1}^n|f(i)|^p\\ &= \sum_{i=1}^{\infty}|f(i)|^p. \end{align*}