If $\int_0^1f^k(x)\;dx\le M$, then $m(\{x\in[0,1]:f(x)>1\})=0$.

Question

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.

Answer 1

$$\int f^k(x)dx = \int f^k(x) 1_{\{y:f(y)\leq 1\}}(x) dx+\int f^k(x) 1_{\{y:f(y)>1\}}(x) dx \leq M $$ From Lebesgue's monotone and bounded convergence theorems we get (applied to second and first integral) $$m(\{y: f(y) = 1\})+\int\lim_{k\to\infty} f^k(x) 1_{\{y: f(y)> 1\}}(x) dx =\\m(\{y: f(y) = 1\})+ \infty\cdot m(\{y: f(y)> 1\})\leq M \implies m(\{y: f(y)\geq 1\}) = 0 $$ Because otherwise the left hand side of the inequality would be infinity. Note that $m(\{y: f(y) = 1\})<\infty$.

Answer 2

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.