TITLE: If $\int_0^1f^k(x)\;dx\le M$, then $m(\{x\in[0,1]:f(x)>1\})=0$.
QUESTION [5 upvotes]: I am having trouble with the following problem from a past analysis qualifying exam:
Let $f(x)$ be a non-negative measurable function defined on $[0,1]$. Suppose there exists a constant $M$ such that
$$\int_0^1f^k(x)\;dx\le M$$ for all $k\ge1$. Prove that $m(\{x\in[0,1]:f(x)>1\})=0$. (Note that $m$ here stands for the Lebesgue measure)
I can't seem to figure out how this bound on the integral helps me. Just following my nose, let $E=\{x\in[0,1]:f(x)>1\}$. Then
$$m(E)=\int\chi_E(x)\;dx<\int f(x)\chi_E(x)\;dx\le\int f^k(x)\chi_E(x)\;dx$$
$$\le\int_0^1 f^k(x)\;dx\le M.$$
Clearly this is not the way to go since I'd like to show that $m(E)=0$. Any help would be very much appreciated.

REPLY [6 votes]: Suppose that $m(E)>0, E=(\{x\in[0,1]:f(x)>1\}$. Let $E_n=m(\{x\in[0,1]:f(x)>1+{1\over n}\})$, $E=\cup_nE_n$ implies that $m(E)=lim_nm(E_n)$ since $E_n\subset E_{n+1}$. This implies that there exists $N$ such that $A=\mu(E_N)>0$.
$\chi(E_N) f^k\leq f^k$  implies that $\int\chi(E_n)f^k\geq A(1+{1\over N})^k\leq M$ for every integer $k$. Contradiction.