Rudin’s RCA theorem $2.24$ Lusin’s theorem

Question

There are the definitions which we need for the proof of the theorem:

There is the theorem:
Suppose $f$ is a complex measurable function on $X$, $\mu(A)$ $\lt$ $\infty$, $f(x)$ $=$ $0$ if $x$ $\notin$ $A$, and $\epsilon$ $\gt$ $0$. Then there exists a $g$ $\in$ $C_c(X)$ such that $\mu$ ({$x:f(x)\neq g(x)$}) $\lt$ $\epsilon$.

Furthermore, we may arrange it so that

$sup$ $x \in X$ $|g(x)|$ $\leq$ $sup$ $x\in X$ $|f(x)|$.

There is the proof:

Assume first that $0$ $\leq$ $f$ $\lt$ $1$ and that $A$ is compact.

Attach a sequence ${s_n}$ to $f$, as in the proof of Theorem $1.17$, and put $t_1$ $=$ $s_1$ and $t_n$ $=$ $s_n$ $-$ $s_{n-1}$ for $n$ $=$ $2$,$3$,$4$,…

Then $2^n$$t_n$ is the characteristic function of set $T_n$ $\subset$ $A$, and

$f(x)$ $=$ $\sum_{n=1}^\infty$ $t_n(x)$ ($x$ $\in$ $X$).

I don't understand why is $2^n$$t_n$ the characteristic function of a set $T_n$ $\subset$ $A$ ?

Any help would be appreciated.

Answer 1

• For $n\ge2$ and for any $y\in[0,n-1),$ $$2^n(\varphi_n-\varphi_{n-1})(y)=\lfloor2^ny\rfloor-2\lfloor2^{n-1}y\rfloor\in\{0,1\},$$ in particular $2^n(\varphi_n-\varphi_{n-1})(0)=0$ and for any $y\in[0,1),$ $2^n(\varphi_n-\varphi_{n-1})(y)\in\{0,1\}.$
  Since $t_n=(\varphi_n-\varphi_{n-1})\circ f$ and $0\le f<1$ and $f=0$ outside $A,$ the result follows.
• The proof for $n=1$ is similar: for any $y\in[0,1),$ $$2\varphi_1(y)=\lfloor2y\rfloor\in\{0,1\},$$ in particular $2\varphi_1(0)=0.$
  Since $t_1=\varphi_1\circ f$ and $0\le f<1$ and $f=0$ outside $A,$ the result follows.