inequalities concerning integration and measure Let $f$ be a non-negative function on $\mathbb{R}^n$ such that $\int_{\mathbb{R}^n} f=1$. Let $p\in(0,1)$. Let $E$ be any measurable subset of $\mathbb{R}^n$. Prove that 
$$
\int _E f^p\leq \mu(E)^{1-p}.
$$
If $\mu(E)<\infty$, prove that 
$$
\int _E \log f\leq -\mu(E)\log \mu(E).
$$
How to prove these? Thanks. 
 A: I don't think that your first inequality is true.
For a counterexample, assume $n = 1$ and consider $f = f_\varepsilon$ for $\varepsilon > 0$ and $f_\varepsilon (x) = \varepsilon \cdot x^{-1+\varepsilon} \cdot \chi_{(0,1)}$. Then indeed $\int f_\varepsilon \, dx = 1$ (with $dx$ denoting Lebesgue measure), but for $\beta \in (0,1)$, your first claim for $E = (0,\beta)$ is equivalent to:
$$
\beta^{\varepsilon}=x^{\varepsilon}\bigg|_{0}^{\beta}=\int_{0}^{\beta}\varepsilon\cdot x^{-1+\varepsilon}\, dx=\int_{E}f\, dx\overset{!}{\leq}\left(\mu\left(E\right)\right)^{1-p}=\beta^{1-p},
$$
i.e. to $1 \leq \beta^{1-p-\varepsilon}$. Because of $\beta \in (0,1)$ this means $1-p-\varepsilon < 0$, i.e. $\varepsilon > 1-p$ which does not hold, e.g. for $\varepsilon = \frac{1}{4}$ and $p=\frac{1}{2}$.
EDIT: For the second inequality, you can use Jensen's inequality. Note that the inequality is trivial for $\mu(E) = 0$, so assume $0 < \mu(E) < \infty$. Then $\frac{\mu}{\mu(E)}$ is a probability measure on $E$ and hence
$$
\exp\left(\int_E \ln(f) \frac{d\mu}{\mu(E)} \right) \leq \int_E f \frac{d\mu}{\mu(E)} \leq \frac{1}{\mu(E)} \cdot \int_{\Bbb{R}^n} f d\mu = \frac{1}{\mu(E)},
$$
which implies
$$
\frac{1}{\mu(E)} \cdot \int_E \ln(f) d\mu \leq \ln\left( \frac{1}{\mu(E)} \right) = -\ln(\mu(E)).
$$
