Lusin's theorem clarification I am looking at lusin's theorem proof in rudin's book and the only detail I cannot understand is when he says that $2^{n}t_{n}$ is the characteristic of some subset $T_{n}$. (3rd line of the proof) clarification will be appreciated. 
 A: In Theorem 1.17 $s_n$ is defined as $s_n=\phi_n\circ f$ for $f\geq 0$ where
$$\phi_n(t)=\begin{cases}
k_{n}(t)2^{-n} \quad\quad 0\leq t< n \\
n \quad\quad n\leq t
\end{cases}$$
and $k_n(t)$ is the unique integer satisfying  $k_n2^{-n}\leq t<(k_n+1)2^{-n}$.
Finally observe that $\phi_{n}\geq \phi_{n-1}$ for all  $n \in \mathbb{N}$ and, 
$$t-2^{-n}<k_n(t)2^{-n}\leq t\quad\quad\ 0\leq t<n\quad\quad *$$
Now, in the proof Rudin first assumes $0\leq f \leq 1$ and it is easily shown that in that case
\begin{align}
t_n=&s_n-s_{n-1}\\
=& k_n(f)2^{-n}-k_{n-1}(f)2^{-(n-1)}
\end{align}
and with (*) we get that 
$$0\leq 2^nt_n=k_n(f)-2k_{n-1}(f)<2$$
but notice that $k_n(f)-2k_{n-1}(f)$ is an integer so $2^nt_n$ only takes the values $0$ and $1$, making it a characteristic function of some set $T_n$; namely $T_n=t_n^{-1}(2^{-n})$
A: Check page 11, section 1.9 of the same book.
It is such function that it is equal 1 for all members of your set $T_n$ and 0 on the complement of the set.
You need the $2^n$ because of the $\delta^{-n}$ in the proof of the theorem 1.17.
