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

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.

Tags:

Real Analysis

Measure Theory

Lebesgue Measure

Lebesgue Integral

Measurable Functions

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