Let ($X,\mathcal{A},\mu$) be a measure space and $f : X \to \mathbb{R}_+ \cup \{\infty\}$ be a non-negative measurable function. Let $\epsilon, p > 0$ and $A = \{x \in X : f(x) \geq \epsilon\}$
I'm trying to prove the inequality $$\mu(A) \leq \frac{1}{\epsilon^p} \int_X f^p \,d\mu $$
If $\mu(A)$ and $\int_X f \,d\mu$ are both $< \infty$, then $\epsilon^{p} \mathbb{1}_{A} \leq f$ and it follows immediately that $$\mu(A) \leq \frac{1}{\epsilon^{p}} \int_X f \,d\mu $$
It also must be that if $\int_X f \,d\mu < \infty$ then $A$ can't have infinite measure. But I am not sure why. Also, I can't continue the proof for $f^p$. Any help would be great.
Best Answer
Let $g=fI_A$. Then $f^{p} \geq g^{p} \geq \epsilon^{p} I_A$. This implies $\int f^{p} d\mu \geq \int \epsilon^{p} I_Ad\mu =\mu (A) \epsilon ^{p}$. Hence $\mu (A) \leq \frac 1 {\epsilon^{p}} \int f^{p}d\mu$. For this argument finiteness of the integrals is not necessary.