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I am having problems with proving the following:

Let $f$ be a $L^p$ function on $[0,1]$, $f:[0,1] \to \overline{\mathbb{R}}$. Prove that

$$\lim_{t \to \infty} t^p \mu(x: |f(x)| \geq t) = 0.$$

I know that the right-hand side value is always finite for each $t \in \mathbb{R}$ due to Chebyshev's inequality. I was also able to prove that

$\lim_{t \to \infty} \mu(x: |f(x)| \geq t) = 0$. But unfortunately can find anything about how fast this value goes to $0$ as $t \to \infty$.

Would be grateful, if you could give an idea and help with this.

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up vote 2 down vote accepted

Sketch of proof:

  • Let $A_k:=\{x\in [0,1], 2^k\leqslant f(x)<2^{k+1}\}$. Since $f\in L^p$, $\sum_{k\in\Bbb Z}2^{kp}\mu(A_k)<\infty$.
  • It's enough to show that for each sequence $\{t_n\}$ of real numbers increasing to $+\infty$, $\lim_{n\to +\infty}t_n^p\mu\{x,|f(x)|\geqslant t_n\}=0$. Indeed, if $t^p\mu(x;|f(x)|\geqslant t)$ doesn't converge to $0$ when $n\to\infty$, we can find $\delta>0$ and a sequence $t_k$ increasing to $\infty$ such that for each $k$, $t_k^p\mu(x;|f(x)|\geqslant t_k)\geqslant \delta$, a contradiction.
  • If $\{t_n\}$ is such a sequence, let $N_n$ such that $2^{N_n}\leqslant t_n<2^{N_n+1}$. Then $$0\leqslant t_n^p\mu\{x,|f(x)|\geqslant t_n\}\leqslant 2^{(N_n+1)p}\mu\{x,|f(x)|\geqslant 2^{N_n}\},$$ and an upper bound is the remainder of a convergent series, namely, $$t_n^p\mu\{x,|f(x)|\geqslant t_n\}\leqslant 2^p\sum_{j\geqslant N_n}2^{jp}\mu(A_j).$$
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