Positive function I am trying to prove the following result:
Let $\mu$ be a Lebesgue measure. Suppose $f$ is a measurable positive application . Show that 
 $\mu\Big(\left\{x\in[0,1]: f>3\right\}\Big)$  is zero
Please help me do do so. Thanks
 A: Call your set $A:=\{f>1\}$. Suppose for contradiction $\mu(A)>0$. Then there is some $p>0$ such that $A_p:=\{f>2^p\}$ has measure $\delta>0$ (this is because you can think of $A$ as the limit of sets $A_p$ as $p\downarrow 0$). 
Then for all $n$, 
\begin{align*}
2^{np} \delta \leq \int_\limits{A_p} f^n \; d\mu \leq \int_\limits{[0,1]} f^n \; d\mu
\end{align*}
and letting $n\rightarrow \infty$ leads to a contradiction.
A: Hint. Supose $\mu([f>1])>0$. Let $k \in \mathbb{N}-\{0\}$ and  $A_k =\left\{x\in[0,1]: f(x)>1+1/k \right\}$. For all $k \in \mathbb{N}-\{0\}$ we have
$$
\int_{[0,1]}f^n(x)d\mu(x)
\geq 
\int_{A_k}f^n(x)d\mu(x)
\geq \int_{A_k}(1+\frac{1}{2k})^n d\mu(x)
\geq (1+\frac{1}{2k})^n\mu(A_k)
$$
The Lesbesgue measure is generated by the finite and open intervals of $\mathbb{R}$. So being a well-behaved measure in the sense that $A_k\uparrow A$ then $\lim_{k\to \infty}\mu(A_k)=\mu(A)$. This means that if $\lim_{k\to\infty}A_k=[f>0]$ then there should be a value $k_0> 0$ such that $\mu(A_{k_0})>0$
