Showing that there exists a sequence $C_n > 0$ such that $ C_n \to ∞, $ with $(C_nf_n)_n$ converges to zero a.e

Question

Suppose that measurable functions $f_n$ on $[0,1]$ converge almost everywhere to zero.

Show that there exist numbers $C_n > 0$ such that $$\lim_{n→∞} C_n = ∞, $$ but the sequence $(C_nf_n)_n$ converges almost everywhere to zero.

Any hint is warmly appreciated. This is an Exercise from the book by Vladimir Bogachev.

Answer 1

As I understand, the question asks about the existence of a sequence of numbers, i.e. $C_n >0$ should be independent of $\omega$. First, we can use Egorov's theorem in order to deduce that the convergence is almost uniform, i.e. for any $\delta >0$ there exists measurable set $A$ with $\mu(A) < \delta$ such that $\sup_{x \notin A} |f_m(x)| \rightarrow 0$.

First, we can choose decreasing sets $A_n$ such that $\lambda(A_n)< 1/n$ and $(f_m)_{m \in \mathbb{N}}$ converges uniformly on $A_n^c$ to the zero function. Note that $N:= \bigcap_{n=1}^\infty A_n$ is a nullset. Next, fix $N(n)$, w.l.o.g. monotone in $n$, such that $|f_m(x)| < 1/n$ for all $m \geq N(n)$ and $x \in A_n^c$. Define $$C_{N(n)} = \sqrt{n}$$ and interpolate, i.e. $C_m = C_{N(n)}$ for all $N(n) \leq m < N(n+1)$.

We have $C_n f_n \rightarrow 0$ $\lambda$-almost eveywhere.

Let $x \in A_n^c$ be fixed and also $\varepsilon >0$. Take $k \geq n$ such that $1/\sqrt{k} < \varepsilon$. For any $m \geq N(k)$ choose $l \geq k$ such that $N(l) \leq m < N(l+1)$, then
$$ |C_m f_m(x)| = \sqrt{l} |f_m(x)| \leq \frac{1}{\sqrt{l}} \leq \frac{1}{\sqrt{k}} < \varepsilon,$$ because $x \in A_n^c \subset A_l^c$.