These are all similar functions. I just need to figure out how they are solved. I will type out all of them, and would like it if someone could give me instructions or a solution to one. Also, if one is not essentially like the rest, I would like to know about that.
$$(a)\ \ \ f(x)=\sum_{n=1}^{\infty}\frac{(-1)^nd_n(x)}{10^n};\text{$d_n(x)$-number of "9"s within the first $n$ digits of decimal notation of $x \in (0,1).$}\\ (b) \ \ \ f(x)=\sum_{n=1}^{\infty}\frac{(-1)^nd_n(x)}{2^n};\text{$d_n(x)$-number of "7"s within the first $n$ digits of decimal notation of $x \in (0,1).$}\\ (c) \ \ \ f(x)=e^{\sum_{n=0}^{\infty}\frac{d_n(x)}{(n+1)!}};\\ \text{$d_n(x)$-represents the $n$-th digit of the decimal notation of $x \in (0,1).$}\\ (d) \ \ \ f(x)=\sum_{n=2}^{\infty}\frac{d_n(x)}{n\log^2n};\\ d_n(x)=\max n_i;\ x=0,n_1n_2n_3...$$ For this last function $d_n(x)=\max n_i$ we proved in class that it is measurable and also that it is almost surely equal to $9$. I would be endlessly grateful for a solution to this (these) problem(s)..
To add to the comment I left, I'll outline a proof of the third function:
first you need to show that $d_n$ is measurable for all $n$, I'll leave that to you (hint: $d_n$ takes on finitely many values for all $n$). Now $g_n:= \sum_{k=1}^n \frac{(-1)^n d_n}{(n+1)!}$ is measurable for all $n$, because it is the sum of measurable functions $(\frac{\pm 1 d_n} {(n+1)!})$ is measurable as well (why?). The sum $\sum_{n=1}^\infty \frac{(-1)^n d_n}{(n+1)!}$ converges for all $x$ in fact uniformly, so $g:=\lim_n g_n=\sum_{n=1}^\infty \frac{(-1)^n d_n}{(n+1)!}$ is measurable because it is the pointwise limit of a sequence of measurable functions. Clearly $e^x$ is measurable (being continuous), so $e^{g(x)}$ is measurable being the composition of measurable functions.