$x=0.x_1 x_2..x_n....$ for some integer $l$ Let $x\in [0,1] $ have the expansion to the base $l$, 
  $x=0.x_1 x_2..x_n....$ for some integer $l$ , the non -terminating expansion being used in cases of ambiguity.
Show that $ f_n(x)=x_n$  is a measurable function of $x$ for each  n.
I have no clear idea regarding this question.  I feels like you need to explicitly write the term $x_n$ as an interval so your $f_n(x)$  lies inside an interval which is measurable. I dont know how to do that.
 A: To show that each function $x_n$ is measurable, we must show that for any $a \in \mathbb{R}$, the set $\{x: f_n(x) > a \}$ is measurable.  For this, we will explicitly write $\{x: f_n(x) > a \}$ as a union of intervals, like you suggested.
We let $n$ be fixed, and we note that if $a < 0$, then $\{x: f_n(x) \geq 0 \} = [0,1]$.  Henceforth, we assume that $a \geq 0$, in which case we have $f_n(x) > a$ if and only if $f_n(x) \geq \lfloor a \rfloor +1$.  This means that we have
$$
\{x: f_n(x) > a \} = \bigcup_{m = \lfloor a \rfloor +1}^{\ell -1} \{x: x_n = m\}. \tag{1}
$$
(Note that if $a > \ell-1$, then the union on the RHS is appropriately a union over an empty index set.)  We must now show that $\{ x: x_n = m\}$ is measurable for $m = \lfloor a \rfloor + 1, 
\dots, \ell -1$.
If $n =1$, then we can write 
$$
\{x: x_1 = m\} = (m \ell^{-1}, (m+1) \ell^{-1}].
$$
(We include the right endpoint and not the left endpoint because of the convention that the non-terminating expansion for $x$ is used in case of ambiguity.)
If $n > 1$, we proceed as follows: we let 
$$
S_n = \Big\{ y \in [0,1] : y = \sum_{k=1}^{n-1}y_k \ell^{-k} \text{ for  some } y_1, \dots , y_{n-1} \in \{0, \dots , \ell-1\}  \Big\}
.$$
This is the set of real numbers in $[0,1]$ we get if we truncate after $n-1$ decimal places in the base $\ell$ expansion.  For each $y \in S_n$, the set $\{ x:x_n = m\}$ contains all $x$ so that $x > y + m \ell^{n-1}$ and 
$ x \leq y+m \ell^{n-1} + \sum_{k \geq n-1}(\ell -1)y^k = y + (m+1)\ell^{n-1}.$  Hence, resuming from (1), we have
$$
\{x: f_n(x) > a \} = \bigcup_{m = \lfloor a \rfloor +1}^{\ell -1} \text{ } \text{ } \bigcup_{y \in S_n} \left( y + m \ell^{-n}, y + (m+1) \ell^{-n} \right],
$$
which implies that $\{x: f_n(x) > a \}$ is measurable.
