Problem. Let $f_n:[0,1]\to [0,\infty)$ be a sequence of measurable functions. Then there exists a sequence $(c_n)_{n=1}^\infty$ of positive reals such that $$ \lim_{n\to \infty}\frac{f_n(x)}{c_n} = 0 $$ for almost all $x\in [0,1]$

We can without loss of generality assume that $f_1\leq f_2\leq f_3\leq \cdots$.

For we can define $g_n=\sup_{k=1}^n f_k$ and obtain a sequence $(c_n)_{n=1}^\infty$ of positive reals such that $\lim_{n\to \infty}g_n(x)/c_n =0$. This same sequence would work for $f_n$'s.

Aslo, the statement is easily proved for increasing sequence of simple measurable functions. For if $s_1\leq s_2\leq s_3\leq \cdots$ is a sequence of measurbale simple functions, then we can define $\alpha_n = \sup_{x\in [0,1]}s_n(x)$ and $c_n=n\alpha_n$ for all $n$. Then clearly $(c_n)_{n=1}^\infty$ works.

I can't see where to go from here.

Can somebody please help?


  • $\begingroup$ Diagonally approximate each $f_n$ by step functions $s_{n,m}$ and pick $c_n = c_{n,n}$. $\endgroup$ – AlexR Mar 11 '15 at 12:41
  • $\begingroup$ @AlexR I don't see how this will lead to the "almost all" clause in the question. Can you please elaborate? $\endgroup$ – caffeinemachine Mar 11 '15 at 12:51
  • $\begingroup$ The almost all comes from the approximation: $$\lim_{m\to\infty} |s_{n,m}(x)-f_n(x)| = 0 \quad\text{a.e.}$$ $\endgroup$ – AlexR Mar 11 '15 at 12:53
  • $\begingroup$ Okay. Let me think about it. Thanks. $\endgroup$ – caffeinemachine Mar 11 '15 at 12:54
  • $\begingroup$ Actually thinking further, you might need a $c_n = c_{n,M_n}$ where $M_n$ assures that $|s_{n,M_n} - f_n|$ is sufficiently small as in $|s_{n,M_n} - f_n| \to 0$ a.e.. $\endgroup$ – AlexR Mar 11 '15 at 12:55

Using Lusin's theorem, you can find for each $n \geq 1$, a compact set $K_n \subseteq [0, 1]$ such that $\mu(K_n) > 1 - 1/n$ the restriction of $f_n$ to $K_n$ is (uniformly) continuous hence bounded by some $M_n$. Now take $c_n = M_n/n$.

  • $\begingroup$ This is a bit of overkill (since this will work in a significantly more general setting with a different proof, and because it takes considerable machinery to prove Lusin's theorem anyway), but it works. +1. $\endgroup$ – Ian Mar 11 '15 at 20:13
  • $\begingroup$ What machinery? Lusin's theorem has a two line proof. $\endgroup$ – Lusin Mar 11 '15 at 21:25
  • $\begingroup$ From Egorov's theorem, sure. Egorov's theorem is more difficult. $\endgroup$ – Ian Mar 11 '15 at 21:42
  • $\begingroup$ On second thought, it is just one line: Delete small measure open sets from the preimages of all rational intervals so that they become relatively open in their union. $\endgroup$ – Lusin Mar 11 '15 at 21:46

For each $n$, there is some real number $R_n$ such that $\lambda \{x\mid f_n(x)\gt R_n\}\lt n^{-2}$. Using the Borel-Cantelli lemma, for almost every $x$, there is an integer $n(x)$ such that for $n\gt n(x)$, we have $f_n(x)\leqslant R_n$. Therefore, we can take $c_n :=R_n^2$.


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