This question is from a past term exercise. I had previsouly asked a question from the same exercises on MSE over here:question
Let $p \in (1, \infty)$. Let $f:X \rightarrow \mathbb{C}$ such that $f \in \mathcal{L}^p(X, \mathcal{M}, \mu)$, and suppose there exists $c>0$ such that $$\mu (\{ x \in X; |f(x)| >t \} ) \leq \frac{c}{t^p}, \forall t>0(*)$$ Let $A \in \mathcal{M}$, and we defined $A_n = \{x\in A; |f(x)| > tn \}$. Show that $$t \sum^{\infty}_{n=0} \mu(A_n) \leq t \mu(A_n) + ck_p t^{1-p} $$ where $k_p$ is a constant that solely depends on $p$.
So my attempt:
$$t \sum^{\infty}_{n=0} \mu(A_n) =t \sum^{\infty}_{n=0} \mu(\{x\in A; |f(x)| > tn \}) \leq t \sum^{\infty}_{n=0} \mu(\{x\in X; |f(x)| > tn \}) $$ And so, by $(*)$, we have:
$$t \sum^{\infty}_{n=0} \mu(\{x\in X; |f(x)| > tn \}) \leq t \sum^{\infty}_{n=0}\frac{c}{(tn)^p} = \frac{tc}{t^p}\sum^{\infty}_{n=0}\frac{1}{n^p} $$
So my choice of the constant would be $k_p = \sum^{\infty}_{n=0}\frac{1}{n^p}$. It seems to me that my proof is complete as normally $t\mu(A) \geq 0$ thus $\frac{tc}{t^p} \leq \frac{tc}{t^p} + t\mu(A)$ , but I feel as if the inequality has to come from my calculations. Any help will be appreciated!
It looks like there's a typo in the problem, or your restatement of it. The correct statement to prove should be:
$$t\sum_{n=0}^\infty \mu(A_n)\leq t\mu(A_0)+ck_pt^{1-p}.$$
The issue is that at $nt=0$ your $(*)$ doesn't hold. So with this in mind:
$$t\sum_{n=0}^\infty \mu(A_n)=t\mu(A_0)+t\sum_{n=1}^\infty\mu(A_n),$$
where the bound for the sum is exactly the same as you've already worked out.