Integrating over a subset (Real Analysis) Let $0 < p < 1$ and $f: X \longrightarrow (0,\infty)$ measuarable with $\int_X f d\mu = 1$.
Prove that $\int_A f^p d\mu \leq \mu(A)^{1-p}$.
Someone please help?
 A: It's a direct application of Hölder's inequality, using the exponents $1/p $ and $1/(1-1/(1/p))=1-p $. So
$$
\int_A f^p\,d\mu\leq\left (\int_A (f^p)^{1/p}\,d\mu\right)^p\left (\int_A1^{1/(1-p)}\,d\mu\right)^{1-p}\leq\mu (A)^{1-p}.
$$
A: This is just an application of Holder's inequality. Try using it to bound
$$\int_Af^pd\mu = \int_{X} f^p \cdot \mathbb{1}_{A} d \mu.$$
Edit: extra clarification.
Holder's inequality states
$\|hg\|_1 \le \|h\|_r\|g\|_s =\left(\int_X |h|^r d\mu\right)^{1/r} \cdot \left(\int_X |g|^s d\mu\right)^{1/s}$
if $r,s \in [1,\infty]$ satisfy $1/r + 1/s =1$.
We are taking $h = f^p$, $g=1_A$ and $r=1/p$ (notice that $r \in [1,\infty]$ since $0<p<1$). To use Holder's we need $1/s = 1 - 1/r = 1-p $ i.e. $s = 1/(1-p)$.
Combining all of this (and noticing that $f$ is positive so we need not worry about absolute values),
\begin{align*}
\int_A f^p d\mu &= \int_X f^p \cdot 1_A d\mu\\
&\le \left(\int_X(f^p)^{1/p}d\mu\right)^p \cdot\left(\int_X (1_A)^{1/(1-p)} d\mu\right)^{1-p}\\
&= 1 \cdot \left(\int_X 1_A d\mu\right)^{1-p}\\
&= \mu(A)^{1-p}
\end{align*}
