Question related to Egorov theorem Let $(X,\mathcal{F},\mu)$ be a finite measure space, and $(f_k)_{k=1}^{\infty}$ be a sequence of measurable real valued functions with
$$\lim_{k\to\infty}f_k(x)=0\quad\mu-a.e.$$
Use Egorov theorem to show that there is a sequence $(t_k)_{k=1}^{\infty}$ of real numbers s.t.
$$\sum_{k=1}^{\infty}|t_k|=\infty,\quad \sum_{k=1}^{\infty}|t_kf_k(x)|<\infty\quad\mu-a.e.$$
Note: It is required to use Egorov's theorem. Any hints are welcomed
 A: I assume that $f_k\downarrow 0$. For a fixed $j$, pick a measurable subset $A_j$ such that $\mu(\Omega\setminus A_j)\lt 2^{-j}$ and $\sup_{x\in A_j}f_k(x)\to 0$ as $k$ goes to infinity. There exists $n_j$ such that $\sup_{x\in A_j}f_{n_j}(x)\leqslant j^{-2}$. We can take $n_j\gt n_{j-1}$. Then define $t_{n_j}=1$ and $t_k=0$ if $k$ is not of the form $n_j$ for some $j$. Since $\{n_j,j\geqslant 1\}$ is infinite, the sequence $(t_i)_{i\geqslant 1}$ takes the value $1$ infinitely many times, hence the series $\sum_{i\geqslant 1}|t_i|$ diverges.      
We use a Borel-Cantelli argument to prove the convergence of $\sum_{k\geqslant 1}t_kf_k$. Indeed, since $\sum_{j=1}^{+\infty}\mu\left(\Omega\setminus A_j\right)\lt +\infty$, then $\mu(\limsup_{j\to\infty}\Omega\setminus A_j)=0$. This means that there is $\Omega'\subset\Omega$ with $\mu(\Omega\setminus \Omega')=0$ and for each $\omega\in\Omega'$, there is $J(\omega)$ such that if $j\geqslant J(\omega)$, then $\omega\in A_j$. For such an $\omega$ we have to prove the convergence of $\sum_{j\geqslant 1}f_{n_j}$. It is enough to sum over the $j\geqslant J(\omega)$ and for these indexes, $0\leqslant f_{n_j}(\omega)\leqslant j^{-2}$.
