I am trying to prove the "extended version" of Markov's inequality:
Suppose $(X, \mathcal S, \mu)$ is a measure space and $h: X \to \mathbb R$ is an $\mathcal S$-measurable function. Prove that
\begin{align}
\mu\left(\left\{ x\in X: \left|h(x) \right|\ge c
\right\}\right) \le
\frac{1}{c^p} \left(\int |h|^p \; d\mu \right)
\end{align}
for any positive numbers $c, p$.
I have the proof for the $p=1$ case. I want to use induction to prove this for any positive number $p$. Of course, the first step in the induction process would be assuming that
\begin{align}
\mu\left(\left\{ x\in X: \left|h(x)\right|\ge c
\right\}\right) \le
\frac{1}{c^p} \left(\int |h|^p \; d\mu \right)
\end{align}
is true. Then, we would need to show that
\begin{align}
\mu\left(\left\{ x\in X: \left|h(x)\right|\ge c
\right\}\right) \le
\frac{1}{c^{p+1}} \left(\int |h|^{p+1} \; d\mu \right)
\end{align}
How can the last inequality be proven given the induction hypothesis? Should I even use induction here?
Best Answer
No need for induction (you also need any $p>0$). Because $$\mu(|f|\geq c )=\mu(|f|^p\geq c^p )=\frac{1}{c^p}\int c^p \mathbb{I}_{\{|f|^p \geq c^p\}}\,d\mu$$ So (because $\mathbb{I}_{\{|f|^p \geq c^p\}} \leq \mathbb{I}_\Omega$) $$\frac{1}{c^p}\int c^p \mathbb{I}_{\{|f|^p \geq c^p\}} \, d\mu \leq \frac{1}{c^p}\int |f|^pd\mu$$ Therefore $$\mu(|f|\geq c ) \leq \frac{1}{c^p}\int |f|^pd\mu$$ $$$$