Is $f(x)=\begin{cases} \ \ \ \ \ \ \ \ \ x,\ \ \ \ x\in C \\ \frac{1}{x}\sin x, \ \ \ \ x\in[0,1]\setminus C\end{cases}$Lebesgue-integrable? Is $$f(x)=\begin{cases} \ \ \ \ \ \ \ \ \ x,\ \ \ \ x\in C \\ \frac{1}{x}\sin x, \ \ \ \  x\in[0,1]\setminus C\end{cases}$$ Lebesgue-integrable?
I honestly don't  know how to prove this. I have the definition that a function $f$ is Lebesgue integrable if there exists a sequence of simple functions $\varphi_n(x)$ such that $\varphi_n(x)< f$ and $\varphi_n \to f, n\to \infty.$
$C$ is indeed, the Cantor set.
 A: The functions $h_1(x) = \chi_C(x)$, $h_2(x) = \chi_{[0,1]\setminus C}(x)$, $g_1(x)=x$, and $g_2(x)=\dfrac 1x \sin x \chi_{\{x \not= 0\}}(x)$ are all Lebesgue measurable.  So is $f(x) = g_1(x)h_1(x) + g_2(x)h_2(x)$.
A: If $g\in C([0,1]),$ then $g$ is the uniform limit of step functions that stay below $g$ on $[0,1].$ And step functions are simple functions. So we can find simple functions $s_n$ such that $s_n(x) \to x$ on $[0,1]$ as above, and similarly we can find simple functions $t_n$ such that $t_n(x) \to (\sin x)/x$ on $[0,1].$ The functions $s_n\cdot \chi_C + t_n\cdot \chi{[0,1]\setminus C},$ which are simple functions, then converge to your $f$ in the desired manner. 
A: The abstract definition of a lesbegue integral via Cantor sets is:
$$ s = \sum_{k}^{n^{2^{n}}}\frac{k+1}{2^{n}} A_{n} $$
So that if you take a relative interval on a cantor set, this can be evaulated as:
$$ s = \sum_{k=1}^{n} f(x_{k}) A_{k} $$
A series of functions less than $\frac{1}{x}\sin{x} $ is:
$$\frac{1}{x}\sin{x} = \lim_{k \rightarrow \infty}\frac{(-1)^{kn}x^{2kn}}{x(2k)!} $$
This is because if a series of functions converges it must be a Cauchy sequence and is therefore under the laws of dominant convergence within the interval.
