Show $f$ is in $L^p$ given Markov-like inequality

I am working on practice problems for a final exam and I'm currently stuck on a question which is stated as follows: Assume $f$ is non-negative and measurable on $\mathbb{R}$ and $m(\{x:|f(x)| \geq \lambda\} \leq \frac{c}{\lambda}$ while $f$ is bounded. Prove that $f$ is in $L^p$ for all $1<p \leq \infty$. Is $\int\limits_{\mathbb{R}} f$ necessarily finite?

My attempt is the following: For the last part of this question, the answer is clearly no because if we let $f(x)=\frac{c}{x}$ when $x>0$ and $f(x)=0$ for $x\leq 0$, then $m(\{x:|f(x)| \geq \lambda\} = \frac{c}{\lambda}$, but $\int\limits_{\mathbb{R}} f= + \infty$.

What I'm struggling with is the first part. Intuitively, I want to argue along the lines of bounding $\int\limits_{\mathbb{R}} f^p$ by $\int\limits_{\mathbb{R}} |\frac{c}{x}|^p$, but I'm not sure if this is necessarily a correct bound. I thought of somehow using Markov/Chebychev's inequality because of the information we're given about the set $\{x:|f(x)| \geq \lambda\}$, but that's led me to a dead end. Any help would be appreciated.

If $f\leq M$ then $\int f^{p} =\int_0^{M} pt^{p-1}m(\{x: f(x)\geq t\})dt \leq \int_0^{M} pt^{p-2}dt<\infty$.

[I have used the formula $\int |f|^{p}=\int_0^{\infty} pt^{p-1}m(\{x: f(x)\geq t\})dt$ which is a consequence of Fubni's Theorem].