This is an exercise of my course of Measure and Integration. I did not know if my ideas are correct, so I will also post my response so they can comment and help me
Let $f(x)$, for $x \geq 0$ such that: $$ f(x)=x^{-1/4}, x>0, f(0)=0 $$ This function is not bounded so is not R-integrable over $[0,1]$. However, its improper integral exists $[0,1]$. Calculate the value of the improper integral $\int_0^1 f$. Now, let the function $g(x)$ over $[0,1]$ such that: $g(x)=0$ if $x\in C(4)$, the Cantor Set, and if $(a,b)$ is one of the open intervals whose meeting produces $[0,1]-C(4)$, so for $x \in (a,b)$, define $$ g(x)= \left\{ \begin{array}{ll} (x-a)^{-1/4} & \textrm{ if} a<x \leq (a+b)/2 \\ (b-x)^{-1/4} & \textrm{ if } (a+b)/2<x<b \\ \end{array} \right. $$ Show that even the improper integral of $ g $ in $ [0,1] $ does not exist. Find the integral of $ g (x) $ on each interval of length $ 4 ^ {-n} $ where $ g $ does not vanish. Remembering that there are exactly $ 2 ^ {n-1} $ of these open intervals of length $ 4 ^ {-n} $, add the values you just calculated and find a value that should be assigned to $\int_0 ^ 1g $.
MY ATTEMPT
(i)
$$ \lim_{n\rightarrow 0^+}{\int_n^1 f(x)dx}=\lim_{n\rightarrow 0^+}{\int_n^1 x^{-1/4}}=\lim_{n\rightarrow 0^+}{\frac{4}{3}(1^{3/4}-n^{3/4})}=\frac{4}{3} $$
(ii)
$$ \lim_{n\rightarrow a+}{\int_n^b g(x)}=\lim_{n\rightarrow a+}{\int_n^b (x-a)^{-1/4}}=\lim_{n\rightarrow a^+}{[(b-a)^{3/4}+(n-a)^{3/4}]}=(b-a)^{3/4}=4^{-3n/4} $$ now, $$ \lim_{n\rightarrow a-}{\int_n^b g(x)}=\lim_{n\rightarrow a-}{\int_n^b (a-x)^{-1/4}}=\lim_{n\rightarrow a^-}{[(a-b)^{3/4}+(a-n)^{3/4}]}= $$ $$ =(a-b)^{3/4}=((-1)(b-a))^{3/4}=(-4)^{-3n/4} \neq 4^{-3n/4} $$ so the improper integral does not exist
(iii) $(2^{n-1}) 4^{-3n/4}=(2^{n-1})(2^{-6n/4})= 2^{(4n-4-6n)/4}=2^{(-2n-4)/4}=2^{(-n-2)/2}$ is the value